Representation theory notes

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REPRESENTATION THEORY NOTES WUSHI GOLDRING

Contents 1. March 19, 2020: Comparing bases, the first orthogonality relation and the character table 2 1.1. Set-up 2 1.2. Recall our two bases of Z(C[G]) 2 1.3. Expressing one basis in the other 3 1.4. The generalized and 1st orthogonality relations 3 1.5. Application I: The inner product on Class(G, C) 4 1.6. Proof of the generalized orthogonality relation 5 1.7. The commutator subgroup and abelianization 5 1.8. Application of the commutator subgroup to degree 1 characters 6 1.9. The character Table 6 2. March 26, 2020: The second orthogonality relation and applications. Pullback and restriction. Symmetric and exterior powers, and their characters. Application of symmetric and exterior powers to nonabelian groups of order 8 and the character table of S5 9 2.1. Restriction 9 2.2. Application of the change of basis 1.3.2 9 2.3. Application of the 1st orthogonality relation II: The 2nd orthogonality relation 10 2.4. Applications of the 2nd orthogonality relation I: Another proof of 2.2.1, following Gorenstein [7] 11 2.5. Interlude: Pulling back representations 11 2.6. Applications of the 2nd orthogonality relation II: Comparing centralizers in G and its quotients 12 2.7. Tensor, symmetric and exterior powers: Characters 13 2.8. Using symmetric and exterior powers to distinguish between groups with the same character table 15 2.9. Using symmetric and exterior powers to complete character tables 16 2.10. Question about the kernel of χ2 for Q8 16 3. April 2, 2020: 17 3.1. The kernel and centralizer of a character 17 3.2. Locating normal subgroups via the character table 19 3.3. Twisting representations by outer automorphisms 20 4. April 9, 2020 20 4.1. Twisting representations by automorphism (continued) 20 4.2. The character table of A5 20 4.3. More about character centralizers 22 4.4. Existence of a faithful irreducible character 22 4.5. Bounding character degrees 23 4.6. More about permutation representations 24 4.7. Other problems from Chapter 2 of [9] 25 4.8. Applications of the 2nd orthogonality relation III : Real elements 25 5. April 16, 2020 26 5.1. Explicit tensor-generation: The Brauer-Burnside theorem 26 5.2. Counting solutions to equations in G as class functions: The Frobenius-Schur Theorem 27 5.3. Involutions: The Alperin-Feit-Thompson Theorem 29 5.4. Algebraic integers 30 6. April 23, 2020 31 6.1. The central character again 31 6.2. Some more preliminaries about algebraic and integral elements 32 6.3. Some of Burnside’s applications of integrality 33 6.4. More integrality: Irreducible character degrees divide the order of the group 34 Date: May 26, 2020. 1

6.5. Applications of the 2nd orthogonality relation IV: Real-valued characters and odd-order groups 7. April 30, 2020 7.1. Reduction of χ(1) divides [G : Z(χ)] to χ(1) divides |G| via tensor products (d’après Tate) 7.2. Induced representations and induced Characters 7.3. Monomial characters and M -groups 8. May 7, 2020. Part I: Frobenius groups and other applications of induction 8.1. More on induced Characters 8.2. Frobenius groups 8.3. Deeper results about Frobenius groups 8.4. Using characters to produce ‘large’ subgroups 9. May 7, 2020. Part II: Every simple group of order 360 is isomorphic to A6 9.1. Review & Motivation 9.2. The local group theory of G simple of order 360 9.3. Simple groups of order 360 = |A6 | I: Via Characters 9.4. Simple groups of order 360 = |A6 | II: A geometric approach 9.5. Connection to Artin L-functions and Artin’s conjecture (not required for the course but very cool) 10. Summary References

35 35 35 37 39 41 41 41 43 43 44 44 45 45 46 46 48 48

1. March 19, 2020: Comparing bases, the first orthogonality relation and the character table This lecture roughly covered (2.12) to (2.17) in Isaacs [9]. The exercise session discussed the commutator subgroup (1.7 (which you can also look up in Dummit & Foote [2]), the character table and some examples of character tables. 1.1. Set-up. Recall our basic set-up: • G is a finite group and we are studying finite-dimensional representations ρ : G → GL(V ) of G over C. (As we said, everything will work just as well over a different algebraically closed field of characteristic zero, such as the subfield of C of algebraic numbers Q.) • We choose representatives ρ1 : G → GL(M1 ), . . . , ρs : G → GL(Ms ) of the s isomorphism classes of irreducible representations of G over C. Recall s is the number of conjugacy classes of G. • Let χj be the character of Mj . We already know that the χj give s distinct class functions on G; the χj form a basis for the space of class functions on G; we write Irr(G) = {χ1 , . . . , χs }.

(1.1.1)

• We know that χj (1) = dim Mj and that s X

χj (1)2 =

j=1

s X

(dim Mj )2 = |G|.

j=1

• We have seen that the character χreg of the regular representation decomposes as χreg =

(1.1.2)

s X

χj (1)χj .

i=1

 1.1.3. Notation differences.

(Dangerous curve ahead!)

(a) Beware that Isaacs uses the notation ρ for the regular character, while we like to use ρ for a representation. (b) Similarly, Isaacs uses the terminology principal character for what I have called the trivial character; this is the linear (=degree 1) character which maps every element of G to 1. 1.2. Recall our two bases of Z(C[G]). Recall that we have two natural bases of the center of the group ring: (a) The basis (ei ) coming from the isotypic decomposition (1.2.1)

C[G] =

s M i=1 2

Mi (C[G]).

Here ei is the ith component of 1 in the decomposition 1.2.1; it is also the identity element of the ring Mi (C[G]). One has 1 = e1 + · · · + es .

(1.2.2)

(b) The basis (Ki ) coming from the conjugacy classes of G. Recall that K1 , . . . Ks denote the conjugacy classes of G and that the X g ∈ Z(C[G]) ⊂ C[G] Ki = g∈Ki

are the sums, in the group ring, of all elements in each conjugacy class. 1.3. Expressing one basis in the other. We want to express the ei in the Ki . Equivalently, we can express the ei in the basis G of C[G] and necessarily the coefficients of conjugate elements of G need to be the same i.e., by grouping together conjugate elements, we get an expression in the basis (Ki ). 1.3.1. Given a class function χ, we write χ(Kj ) for the common value χ(g) for all g ∈ Kj . Note that two elements g, h ∈ G are conjugate in G if and only if their inverses g −1 , h−1 are conjugate (if xgx−1 = h, then xg −1 x−1 = (xgx−1 )−1 = h−1 ; this is a consequence of "conjugation by x" being an automorphism). Therefore Kj−1 = {g −1 |g ∈ Kj } is again a conjugacy class and χi (Kj−1 ) makes sense. 1.3.2. Theorem (Change of basis, [9], (2.12)). One has 1 X χi (1)χi (g −1 )g. (1.3.3) ei = |G| g∈G

Equivalently, s

ei =

(1.3.4)

1 X χi (1)χi (Kj−1 )Kj . |G| j=1

1.3.5. Proof. The idea is to use what we know about the regular character χreg . Write ei =

X

ag g ∈ C[G]

g∈G

in the basis G of C[G]. We want to determine the coefficients ei . One has (1.3.6)

χreg (ei g −1 ) = χreg (

X

ah hg −1 ) =

h∈G

X

ah χreg (hg −1 ),

h∈G

where the second equality uses that a character is a linear transformation C[G] → C. By our computation of χreg , the only potentially nonzero term in (1.3.6) is the one involving the identity (the h = g term), namely ag χreg (gg −1 ) = ag χreg (1). Since χreg (1) = |G|, we conclude that (1.3.7)

χreg (ei g −1 ) = ag |G|.

We now plug in what we found (1.3.7) into the decomposition (1.1.2) of the regular character; this gives (1.3.8)

ag |G| = χreg (ei g −1 ) =

s X

χj (1)χj (ei g −1 ).

j=1

Since ρj (ei ) = δij I and since ρj : C[G] → End(Mj ) is an algebra homomorphism, one has  0 if i 6= j −1 −1 −1 ρj (ei g ) = ρj (ei )ρj (g ) = δij ρj (g ) = ρi (g −1 ) if i = j Hence only the i = j terms survives in the decomposition into irreducible characters (1.3.8), so ag |G| = χi (1)χi (g −1 ). 

This gives (1.3.3). 1.4. The generalized and 1st orthogonality relations. 1.4.1. Lemma. The ei are pairwise orthogonal, central idempotents in C[G]: (1.4.2)

ei ej = δij ei 3

1.4.3. Proof. We already know that the ei are central, i.e., that they lie in the center Z(C[G]). Since the ideals (isotypic components) Mi (C[G]) annihilate either other, i.e., Mi (C[G])Mj (C[G]) = 0 for i 6= j, we get ei ej = ej ei = 0 for all i 6= j; this is the orthogonality. Then using the decomposition of 1 as sum of the ei (1.2.2) gives 1ei = ei ei = e2i , so the ei are idempotents.  1.4.4. When we defined types of idempotents e, we also had the notions of primitive idempotent and central pimitive idempotent, which mean that e is not a sum of two commuting idempotents and that e is not a sum of two idempotents which both lie in the center, respectively. 1.4.5. Question. Are the ei primitive in C[G]? Are they central primitive? We now interpret (1.4.2) in terms of the other basis of Kj ; this gives us the 1.4.6. Theorem (Generalized orthogonality relation). For all h ∈ G, one has   1 X χi (h) χj (h) (1.4.7) χi (gh)χj (g −1 ) = δij = δij |G| χi (1) χj (1) g∈G

Before proving the theorem, let us record the important corollary gotten by setting h = 1 in (1.4.7): 1.4.8. Corollary (1st orthogonality relation). For every pair of irreducible characters χi , χj ∈ Irr(G), one has: 1 X χi (g)χj (g −1 ) = δij (1.4.9) |G| g∈G

Equivalently, in terms of the conjugacy classes K1 , . . . , Ks : s 1 X |Kl |χi (Kl )χj (Kl −1 ) = δij (1.4.10) |G| l=1

1.5. Application I: The inner product on Class(G, C). 1.5.1. Definition. We define an inner product (=positive definite Hermitian form) on the space of class functions [·, ·] : Class(G, C) × Class(G, C) → C by [χ, ψ] =

1 X χ(g)ψ(g). |G| g∈G

1.5.2. Explanation: Hermitian. Hermitian means that [, ] is additive in both variables, linear in the first variable but conjugate linear in the second: [aχ, ψ] = a[χ, ψ] and [χ, bψ] = b[χ, ψ] for all a, b ∈ C. Hermitian also means conjugate-symmetric: [χ, ψ] = [ψ, χ]. 1.5.3. Explanation: Positive definite. Positive definite means that [χ, χ] is real and ≥ 0 for all χ, and [χ, χ] = 0 if and only if χ = 0. (The ≥ 0 part is often called "positive semi-definite".) P The positive definiteness is seen by writing χ in the basis of the irreducible characters: If χ = ai χi , then X 2 [χ, χ] = |ai | , where |a is the usual absolute value, so that |a|2 = aa for all a ∈ C. Since χ(g −1 ) = χ(g) for every character χ (but not for every class function), the first orthogonality relation can be restated as: 1.5.4. Corollary. The set Irr(G) of irreducible characters is an orthonormal basis for Class(G, C) equipped with the inner product [, ]. 1.5.5. Multiplicities. Recall that given two representation V, W , we have the multiplicity nV (W ) which measures the number of times V appears in W . Given two characters χ, ψ we can also define the multiplicity nχ (ψ). We can do this in two equivalent ways: (a) We choose V, W having characters χ, ψ respectively and define nχ (ψ) := nV (W ). This is independent of the choice of V, W . Why? Ps (b) We let nχ (ψ) count Ps the number of times the expression χ = i=1 ai χi for χ in irreducibles appears in the expression ψ = i=1 bi χi for ψ in irreducibles. That is we take the minimum of bbi /ai c over the i with ai 6= 0. Among other things, we now obtain the irreducibility criterion which was advertised on the first homework. 1.5.6. Corollary. Let χ, ψ be two characters of G. Then 4

(a) The pairing [χ, ψ] = [ψ, χ] is symmetric and always a non-negative integer (when restricted to characters). (b) The character χ is irreducible if and only if its norm (squared) [χ, χ] = 1. (c) If χ is irreducible, then [χ, ψ] = nχ (ψ) is the number of times χ appears in ψ. 1.5.7. Proof. All written inPthe basis of irreducible Pof the statements follow from the fact that when a character isP characters χ = ai χi , the coefficients ai are integers. In particular ai = ai and |ai |2 = a2i is a sum of squares 2 of integers (possibly involving terms 0 = 0).   1.5.8. If χ is just a class function, it could happen that [χ, χ] = 1 but that χ is not a character. For example, this happens if χ = ζχi and |ζ| = 1 but ζ 6= 1.  1.5.9. If χ is not irreducible, then [χ, ψ] 6= 0 does not imply that χ appears in ψ: For example, we could have χ = χ1 + 2χ2 and ψ = 3χ2 + 2χ3 . Then [χ, ψ] = 2 · 3 = 6 but neither one appears in the other; the nonzero inner product reflects that some irreducible character (namely χ2 in this example) appears in both. 1.6. Proof of the generalized orthogonality relation. 1.7. The commutator subgroup and abelianization. 1.7.1. Commutators. Recall that the commutator of two elements x, y in a group G is [x.y] = xyx−1 y −1 Thus the commutator is trivial [x, y] = 1 if and only if x, y commute. The conjugate of a commutator is again a commutator: For all g, x, y ∈ G, using gx−1 g −1 = (gxg −1 )−1 as in 1.3.1 and inserting g −1 g everywhere gives (1.7.2)

g[x, y]g −1 = [gxg −1 , gyg −1 ].

 The product of two commutators need not be a commutator. Can you find an example? 1.7.3. The commutator subgroup. We define the commutator subgroup G0 of G to be the subgroup generated by all the commutators. Thus elements of G0 are finite products of commutators. Other notation for G0 is [G, G]; it sometimes (especially in the theory of matrix and algebraic groups) also called the derived subgroup and denoted Gder . The commutator subgroup is in many ways opposite to the center: For example G is abelian if and only if the G = Z(G) (center as large as possible) if and only if G0 = {1} (commutator subgroup as small as possible. 1.7.4. Characterization of the commutator subgroup. The commutator subgroup G0 is characterized as the smallest normal subgroup of G such that the quotient G/G0 is abelian. We explain why: (a) We see that G0 is normal in G by (1.7.2). (b) Recall that a subgroup H of G is characteristic in G if H is stable under all automorphisms of G. Then G0 is also characteristic in G, because ϕ([x, y]) = [ϕ(x), ϕ(y)] for all automorphisms ϕ ∈ Aut(G) (note that (1.7.2) is the special where ϕ is the inner automorphism "conjugation by g"). (c) The quotient is abelian because [π(x), π(y)] = π([x, y]) = 0, where π : G → G/G0 is the projection (and recall that π is surjective, so every α ∈ G/G0 has the form π(x) for some x ∈ G). (d) Minimality: Assume N is normal in G and that G/N is abelian. We claim that G0 ⊂ N ; by the isomorphism theorem, this is equivalent to the statement that the projection πN : G → G/N factors through the projection π 0 : G → G/G0 . Since G/N is abelian, πN ([x, y]) = [πN (x), πN (y)]0 for all x, y ∈ G. So all commutators, and hence all of the subgroup G0 which they generate, lie in the kernel N of πN . 1.7.5. Abelianization. The group G/G0 is therefore the largest abelian quotient of G. For this reason G/G0 is called the abelianization of G or "G made abelian" and One often writes Gab := G/G0 . 1.7.6. Example. (a) For n > 2, the commutator subgroup of Sn is Sn0 = An . (b) If G is nonabelian simple, then G0 = G. (c) If G is dihedral D8 or quaternion Q8 of order 8, then G0 = Z(G) of order 2, since G is nonabelian but every group of order 4 = 22 is abelian. (d) If you have seen free groups, then the free group on d generators Freed has abelianization Freed /Free0d ∼ = Zd 0 free abelian of rank d. It is also interesting that Free2 is free on countably infinitely many generators, illustrating that a finitely generated nonabelian group may have a non-finitely generated subgroup. (e) [Not needed for the course] If you like topology and have seen homology/cohomology and/or the fundamental group, then note that G/G0 = H1 (G, Z) and that given a reasonable topological space X, one has π1 (X)ab = H1 (X, Z). 5

1.7.7. Perfect groups. A group G is called perfect if G0 = G. The group SL(2, F5 ) is the smallest example of a perfect group which is not simple (it has center {±I} of order 2 and that is its only normal subgroup, the quotient is isomorphic to A5 ). 1.7.8. The abelianization as a reflector (not needed for the course). For those who have seen categories, G 7→ Gab = G/G0 is a functor from the category of groups to that of abelian groups which is an example of a reflector in the sense of Freyd [3, Chap. 3, Exercise F]. 1.8. Application of the commutator subgroup to degree 1 characters. The following generalizes what we have already seen, that an abelian group has as many irreducible representations as its order, as all the irreducible representations are of dimension 1. 1.8.1. Lemma. The number of degree 1 characters of G over C is the order of the abelianization G/G0 . 1.8.2. Proof. Assume χ : G → C× is a homomorphism (=degree 1 character of G). Then the image of χ is abelian. Hence χ factors through G/G0 . So there are at most |G/G0 | degree 1 characters of G. Conversely, every one of the |G/G0 | characters of G/G0 can be composed with the projection π 0 : G → G/G0 to give |G/G0 | distinct degree 1 characters of G.  1.9. The character Table. Let G be a finite group with precisely s conjugacy classes. The character table of G is an s × s matrix which records the values of the s irreducible complex characters of G (which traditionally label the rows) on the s conjugacy classes (which traditionally label the columns). It is probably easiest to first look at examples below and then come back here to the "usual manual" or to go back-and-forth between the two as one does when building furniture etc. The character table of G is really only well-defined up to permutations of rows and columns: There is in general no canonical way to order neither the conjugacy classes of G nor its irreducible characters. If we choose an orderings K1 , . . . , Ks and χ1 , . . . , χs of the conjugacy classes and irreducible characters respectively, then the (i, j) entry of the character table is χi (Kj ). 1.9.1. Usual Manual. It is customary to let χ1 = 1 be in the first row. It is also common to have the following "metadata" recorded close to the character table (above, to the left, below etc. ): (a) (b) (c) (d)

Write the name of each character to the left of the row which contains its values Write a name or representative of each conjugacy class above Record additional information which is useful for making computations in the table, most importantly Record the size of each conjugacy class.

Dummit & Foote [2] and Isaacs [9] both have an additional "metadata row" right below the names of the conjugacy classes, so that the size of each class appears right below it. Instead, I will write (a)

g(b) , where g ∈ Ki represents the conjugacy class Ki , the integer a is the common order |g| of g and all of its conjugates, and b = |Ki | is the size of the conjugacy class. As we shall see, for example in computing characters of symmetric and exterior powers (see 2.7.6), it is often necessary to also know where each class maps to under powers, i.e., , the class of g 2 , g 3 , . . .. Writing the order of the elements gives us some indication of this: If |g| = 4, then |g 2 | = 2 (but of course there could be multiple classes of order 2 and of order 4 and we will have to be careful to understand which maps to which under squaring x 7→ x2 . 1.9.2. Additional "metadata" rows. I like to write a double or squigly line under the last row of the character table and then I add additional rows to record computations. For example, computing a character table (i.e., finding all the irreducible characters explicitly) will often involve computing several characters which are not irreducible and breaking those up into irreducibles. For example, if we know an irreducible character χ, then we can compute its square χ2 (which is the character of the tensor product of a representation which affords χ, see 2.7, esp. Lemma 2.7.2), and record that at the bottom in a metadata row. We can then pair [χ2 , χi ] with all the irreducible characters χi which we already know, and if we know enough χi then we will be able to decompose χ2 . In many ways the best situation is if we can show that χ2 is the sum of several irreducible characters we already know and P one other irreducible character we had not yet found. We can then conclude that a difference χ2 − ai χi is a ‘new’ irreducible character (by showing that it pairs with itself to 1 after we have shown it to be a character). We then add the new irreducible character to the main character table and continue. Again it is probably easier to understand all of this in examples. 6

1.9.3. Example I: Character table of a cyclic group. Assume G = hxi is cyclic of order n, generated by some x ∈ G. Choose a primitive nth root of unity ζ ∈ µprim . For every 0 ≤ j ≤ n − 1, there is a unique linear character χj n determined by χj (x) = ζ j ; this gives n irreducible linear characters and thus determines the whole character table; the trivial character corresponds to j = 0. To get used to the notation, the table for n = 4 is given in Table 1.9.3, where ζ = i is a primitive 4 root of unity and i2 = −1 as usual. We have added the regular P character χreg as a metadata row. As a partial check of our computation, we can verify the relation χreg = χ∈Irr(G) χ(1)χ which we proved before (1.1.2). We can also check cases of the 1st orthogonality relation, such as [χ2 , χ3 ] = 0 and [χ2 , χ2 ] = 1. Table 1.9.4. Character table of Z/4 with regular character added (1)

(4)

(2)

(4)

Z/4

1(1)

x(1)

(x2 )(1)

(x3 )(1)

χ0 = 1

1

1

1

1

χ1

1

i

−1

−i

χ2

1

−1

1

−1

χ3

1

−i

−1

i

χreg

4

0

0

0

1.9.5. Example II: Character table of an abelian group. If G is abelian but not cyclic, then write G as a product of cyclic groups G∼ = Z/n1 × · · · × Z/nd . prim Let ζi ∈ µni be a primitive ni th root of unity. Then for each i we have the "building block" characters χi,j of Z/ni defined by χi,j (ζi ) = ζij . We can view each χi,j as a character of G by setting it to be one on all the other factors (this is a special case of "pulling back representations", see 2.5.1). Then any sequence (j1 , . . . , jd ) ∈ Z/n1 × · · · × Z/nd gives the product character χ1,j1 · · · χd,jd . So we have constructed n1 · · · nd = |G| distinct linear characters; hence these make up the entire character table of G. 1.9.6. Remark. For any group G (not necessarily abelian, not necessarily finite), the set of linear characters of G forms a group, often denoted X ∗ (G). Multiplication is defined pointwise (χψ)(g) = χ(g)ψ(g) for χ, ψ ∈ X ∗ (G). When G is finite abelian, the map (j1 , . . . jd ) 7→ χ1,j1 · · · χd,jd is a group isomorphism ∼

G → X ∗ (G). When G is not abelian, the group X ∗ (G) does not give good information about G, e.g., if G = G0 is perfect (1.7.7, in particular if G is nonabelian simple) then X ∗ (G) = {1}. 1.9.7. Example III: Character table of S3 . Let G = S3 , the smallest nonabelian group (recall all nonabelian groups of order 6 are isomorphic to S3 ∼ = D6 ). The commutator subgroup is S30 = A3 , so there are two linear characters: The trivial one 1 and the sign character sgn given by  1 if σ is even (1.9.8) sgn(σ) = −1 if σ is odd We also know that the regular character χreg is given by  |S3 | = 6 χreg (σ) = 0

if if

σ=1 σ 6= 1

This much gives us the incomplete Table 1.9.9 with metadata row given by χreg . (We call the missing character χ2 because we will see in a moment that its degree is 2.) There are several ways to find the missing row. Here are three: 7

Table 1.9.9. Incomplete character table of S3 with regular character added (1)

(2)

(3)

S3

1(1)

(12)(3)

(123)(2)

1

1

1

1

sgn

1

−1

1

6

0

0

χ2 χreg

1.9.10. Method 1. Since only one row is missing, we can solve for it using the first orthogonality relation. Using X χ(1)2 = |G| χ∈Irr(G)

gives us χ2 (1)2 = 6 − 1 − 1 = 4, so χ2 (1) = 2. Let χ2 ((12)) = b and χ2 ((123) = c. Then 0 = [χ2 , 1] = 2 · 1 + 3b + 2c and 0 = [χ2 , sgn] = 2 · 1 − 3b + 2c. Adding the two relations gives c = −1. Then we solve for b and find b = 0. So the completed table is: Table 1.9.11. Completed character table of S3 with regular character added (1)

(2)

(3)

S3

1(1)

(12)(3)

(123)(2)

1

1

1

1

sgn

1

−1

1

χ2

2

0

-1

χreg

6

0

0

Note that instead of using the sums of squares identity, we could have instead written χ2 (1) = a and then used that [χ2 , χ2 ] = 1 gives a third relation. So we have three relations: a + 3b + 2c a − 3b + 2c |a|2 + 3|b|2 + 2|c|2

=0 =0 =6

Adding the first two gives a = −2c. Substituting into the third gives 6|c|2 = 6. Hence |c| = 1. We can conclude that c = −1 since a is an integer > 0 (as the dimension of a representation). 1.9.12. Method 2. Recall from Homework 1 the character (1.9.13)

Std(σ) = (number of fixed points of σ) − 1

of Sn . We shall henceforth call Std the standard character of Sn . We have seen that Std is irreducible for all n (using that Sn acts doubly-transitively on {1, . . . , n}. For S3 , one has χ2 = Std. 1.9.14. Method 3. We know that [χreg , 1] = χreg , sgn] = 1 since the multiplicity of an irreducible character in the regular character is its dimension (also we can compute it using the table, which is why the metadata row is useful). Hence the difference ψ := χreg − 1 − sgn is a character. We compute [ψ, ψ] = 4; it follows that ψ is twice an irreducible character and we obtain χ2 (σ) = ψ(σ)/2 (if we hadn’t yet computed χ2 by a previous method). 8

2. March 26, 2020: The second orthogonality relation and applications. Pullback and restriction. Symmetric and exterior powers, and their characters. Application of symmetric and exterior powers to nonabelian groups of order 8 and the character table of S5 2.1. Restriction. Given a representation ρ : G → GL(V ) and a subgroup H of G, we can restrict ρ to H; it gives a representation ρ|H of H. I like to write ResG H ρ for the restriction, since it seems to me to be descriptive and functorial notation. Isaacs writes ρH for the restriction. As we did for the multiplicities nV (W ), anything that is defined for representations and only depends on the isomorphism class can be defined for characters (of finite groups over C) given that we know that equality of characters determines isomorphism of representations. So, given a character χ of G, we also have χH = ResG H χ, the restriction of χ to H. As we did for the multiplicities, the restriction of a character can also be defined directly by restricting χ, as a function on G, to H. Again the two definitions agree. 2.1.1. Restriction functor (not required for the course). If you have seen a bit about categories and functors, then one can look at RepC (G), the category of (finite-dimensional) representations of G over C, and then restriction is a functor (2.1.2)

ResG H : RepC (G) → RepC (H).

If you haven’t seen categories before, the category RepC (G) is just a way of keeping track of all the representations of G and also all of the representation maps between them, i.e., the G-equivariant maps between representations. The saying that restriction is a functor is saying that not only does it map one representation of G to one of H, but it also respects maps between representations (so if there is a G-equivariant map between ρ and σ, we can restrict G the map to an H-equivariant map between ResG H ρ and ResH σ.  2.1.3. The restriction of an irreducible representation need not be irreducible. For example, we have discusssed that the tautological representation Id : GL(n) → GL(n) is irreducible (because GL(n, k) acts transitively on k n \{0}), but we have also seen that the restriction to the subgroup H of permutation matrices (H ∼ = Sn ) is reducible (contains the trivial representation). 2.1.4. Branching. It is one of the oldest and most-studied problems in representation theory to understand how an irreducible representation of G decomposes when restricted to a subgroup H. This problems is known as branching. Some basic branching theorems for compact Lie groups are discussed in Knapp [10, Chap. 9]. One often considers a sequence of groups indexed by a positive integer n and studies "branching in the family". For example, one can consider branching from GL(n) to GL(n − 1), where we identify GL(n − 1) with the subgroup of GL(n) of block diagonal matrices have 1 in the top left and then A ∈ GL(n − 1) in the bottom right (n − 1) × (n − 1) block. It is then a "branching theorem" that when an irreducible representation of GL(n) is restricted to GL(n − 1), every irreducible representation of GL(n − 1) which appears in the decomposition does so with multiplicity one. For an example of a sequence of groups where studying branching "uniformly in the family" is interesting, one can consider all the symmetric groups Sn , with Sn embedded in Sn+1 as the subgroup fixing n + 1 and so on. Then one has the following multiplicity free branching theorem: 2.1.5. Theorem ([1], Th. 40.1, Cor. 40.1). The irreducible representations of Sn are parameterized by partitions λ of n. There is an explicit combinatorial rule which says when the irreducible representation of Sn−1 corresponding to a partition µ of n − 1 appears in the restriction to Sn−1 of an irreducible representation of Sn parameterized by a partition λ of n. In particular, the rule implies that the restriction of every irreducible representation of Sn to Sn−1 is multiplicity-free. 2.2. Application of the change of basis 1.3.2. The following shows that "characters separate points" where the "points" are the conjugacy classes. We know that conjugate elements of a group G are indiscernible by characters: Every character takes the same value on conjugate elements. The following shows that, more surprisingly the converse is also true, i.e., characters suffice to detect conjugacy: 2.2.1. Theorem. Let x, y ∈ G. Then x and y are conjugate in G if and only if χ(x) = χ(y) for all χ ∈ Irr(G). 2.2.2. Proof 1. A second proof using the 2nd orthogonality relation is given later in 2.4. We have already seen that any character is constant on conjugacy classes. So assume χ(x) = χ(y) for all χ ∈ Irr(G). We want to show x, y are conjugate in G. Since every P χ is constant on conjugacy classes, one has χ(Ki ) = |Ki |χ(Ki ) for every conjugacy class Ki (recall that Ki = x∈Ki x ∈ C[G]). Thus, if x ∈ Ki and y ∈ Kj , then χ(Ki ) = χ(Kj ) for all χ ∈ Irr(G). We have seen that {ei }si=1 and {Ki }si=1 are two bases of Z(C[G]). Therefore we can express s s X X bl e l . (2.2.3) Ki = al el and Kj = l=1

l=1 9

To find the coefficients al , bl amounts to inverting the matrix which we get from (1.3.4) which expressed ei explicitly in the basis {Kj }sj=1 . However, for the proof we do not need to know the al , bl explicitly. Note that x, y are conjugate in G if and only if Ki = Kj , which in turn is if and only if al = bl because {ei }si=1 is a basis. Write Irr(G) = {χ1 , . . . χs } and recall that χi (ej ) = δij χi (1). Hence χl (Ki ) = al χl (1) and χl (Kj ) = bl χl (1) (when χl is applied to the sums (2.2.3), only the l-term survives). Since χl (Ki ) = χl (Kj ), we conclude that al χl (1) = bl χl (1), so al = bl as desired. (Again we have a priori used that we are in characteristic zero to cancel χl (1) on both sides.)  2.3. Application of the 1st orthogonality relation II: The 2nd orthogonality relation. 2.3.1. Idea. As with the expression 1.3.2 of the ei in the basis Kj , in the first orthogonality relation 1.4.8 we have a sum over g ∈ G but the coefficients of conjugate elements g, h ∈ G are equal, so it is also a sum over conjugacy classes as in (1.3.4). The idea of the second orthogonality relation is in two parts: (a) Rewrite the first orthogonality relation as a matrix equation. (b) Perform a transpose to the equation: Instead summing χi (Kj ) over conjugacy classes (summing over j), we will be summing χi (Kj ) over irreducible characters (summing over i). 2.3.2. Theorem. For all x, y ∈ G, one has (2.3.3)

X

χ(x)χ(y

−1

 ) = δxy | CentG (x)| =

χ∈Irr(G)

| CentG (x)| 0

if if

x, y conjugate in G . x, y not conjugate in G

Since χ(y −1 ) = χ(y) for every character χ, we again have the alternate formulation:  X | CentG (x)| if x, y conjugate in G (2.3.4) χ(x)χ(y) = δxy | CentG (x)| = . 0 if x, y not conjugate in G χ∈Irr(G)

2.3.5. Proof. Consider the matrices X := (χi (Kj ))i,j and D := diag(|K1 |, . . . , |Ks |) = (δij |Ki |)i,j in Ms (C). Since XDt X =

X

χi (Kν ) |Kν | χj (Kν ),

ν

the 1st orthogonality relation in terms of conjugacy classes 1.4.10 is equivalent to the matrix equation |G|I = XDt X. Since |G|I is in the center of Ms (C), multiplying by X −1 on the left, moving it over and then multiplying back by X on the right gives |G|I = Dt XX. Unpacking the modified matrix equation back into a sequence of linear equations gives X |G|δij = |Ki | χν (Ki )χν (Kj ). ν

We conclude because | CentG (Ki )| = |G|/|Ki | (orbit-stabilizer). 2.3.6. Notation. In analogy with 1.5.1, for x, y ∈ C[G] we can define X (2.3.7) [x, y] = χ(x)χ(y). χ∈Irr(G)

We can write [x, y]Irr(G) and [χ, ψ]G when we want to distinguish between the pairings of elements over characters and those of characters over elements respectively. The pairing [, ]Irr(G) : C[G] × C[G] → C will again be bi-additive and Hermitian, because a character χ : C[G] → C is linear (because the trace tr : Mn (C) → C is linear). If we restrict to [, ]Irr(G) : Z(C[G]) × Z(C[G]) → C, then [x, y]Irr(G) is also positive definite: P P P If x = ai Ki , then [x, x] = |ai |2 |Ki |2 | Cent(Ki )| = |G| |ai |2 |Ki |.  10

 2.3.8. If x 6∈ Z(C[G]), then it could be that [x, x] = 0. For example, if g, h ∈ G are distinct but conjugate, then consider x = g − h ∈ C[G] \ Z(C[G]). Then [g, h] = [g, g] so that [x, x] = [g − h, g − h] = 2[g, g] − 2[g, g] = 0. 2.3.9. Remark. The more algebraic formulation 2.3.3 is more useful if we want to apply algebra, or if we want a statement which makes sense more generally (e.g., over any algebraically closed field of characteristic 0). The formulation using complex conjugation (2.3.4) is more useful when we want do positivity arguments e.g., argue that some number, is nonzero because it is real and strictly positive (like the absolute value |z| of a nonzero complex number z). We see the latter strategy employed in 2.4 and 2.6. 2.4. Applications of the 2nd orthogonality relation I: Another proof of 2.2.1, following Gorenstein [7]. Again the interesting direction is to assume that χ(x) = χ(y) for all χ ∈ Irr(G) and to show that x, y are conjugate in G. Our assumption implies that χ(x)χ(y) = χ(x)χ(x) = |χ(x)|2 ≥ 0 for every χ ∈ Irr(G). Moreover, when χ = 1 is the trivial character, 1(x)1(y) = 1(x)1(y) = 1 · 1 = 1 > 0. Hence X

X

χ(x)χ(y) =

χ∈Irr(G)

χ(x)χ(x) > 0.

χ∈Irr(G)

By the 2nd orthogonality relation x, y are conjugate in G.



2.5. Interlude: Pulling back representations. Before proceeding to our next application 2.6.1 of the 2nd orthogonality relation, we pause to observe the relationship between the irreducible representations/characters of G and those of its quotients G/N . 2.5.1. Pullback. More generally, assume ϕ:G→H is a homomorphism of groups (ϕ will be of the form G → G/N for some N if and only if ϕ is surjective). Given a representation ρ : H → GL(V ) of H the pullback of ρ to G, denoted ϕ∗ ρ is the composition G

ϕ

/H

ρ

/ GL(V ); 4

ρ◦ϕ=ϕ∗ ρ

the pullback is a representation of G. Note that restriction 2.1 is a special case of pullback. Namely when i : H → G is the inclusion of a subgroup H in G and r : G → GL(W ) is a representation of W , then i∗ r = ResG H r.  2.5.2. Note that the order of H has been reversed: At first it was the target of ϕ but in the example of a subgroup H is the subgroup and G is target of the inclusion i (this is simply because I like to denote both a "generic" subgroup and target group by H when both are not used simultaneously).  2.5.3. While the category theory in the next two points is not needed for the course, it is important to understand the functions involved and the (non-)reversal of the order; to gain intuition for this one may pretend that things like Rep(G) are sets and think of ϕ∗ as a usual function between sets. 2.5.4. Remark. (Not needed for the course) As with restriction, pullback is also a functor ϕ∗ : RepC (H) → RepC (G). If you have not seen the "upper-star" notation in a different situation, it is used to tell us that pullback is a contravariant operation i.e., that the order between G and H is inverted between ϕ : G → H and ϕ∗ : Rep(H) → Rep(G). We could say that pullback is a contravariant functor when viewed as a functor from the category of groups to the (2-)category whose objects are the categories Rep(G) (or to a larger. 11

 2.5.5. (Not needed for the course.) As a functor from the category Rep(H) to Rep(G), ϕ∗ is covariant in that an H-equivariant linear map a : V1 → V2 between representations of H induces ϕ∗ (a) : ϕ∗ V1 → ϕ∗ V2 and the order of V1 , V2 is the same. So one has to be very careful with "covariance" and "contravariance" to identify what is the variable (e.g., the group or its representation) whose order is changing or staying the same. 2.5.6. In general, an operation involving pre-composing (i.e., f ◦ · · · ) with a function f will be contravariant and it will be natural to call it f ∗ ; an operation involving post-composing (i.e., · · · ◦ f will be covariant and it will be natural to denote it f∗ . In our situation we can see the covariance and contravariance depending on whether we take f = ϕ (contravariant) or f = a (covoriant) in the following diagram: G

ϕ

/H ρ ϕ∗ ρ

a∗ ρ

( # GL(V1 )

a

)

/ GL(V2 ).

2.5.7. Lemma. Let ϕ : G → H be a homomorphism of groups. Let ρ : G → GL(V ) and r : H → GL(W ) be representations of G, H respectively. Then: (a) If Z ⊂ W is an H-stable subspace (for the action by r), then Z is also G stable (for the action by the pullback ϕ∗ r). In particular, if (W, r) is an irreducible representation of H, then (ϕ∗ r, W ) is and irreducible representation of G. (b) If ker ϕ ⊂ ker ρ, then ρ factors through ϕ. 2.5.8. Remark. Note that ρ factors through ϕ if and only if ρ is the pullback of some representation of H. Thus 2.5.7(b) amounts to determining the image of ϕ∗ : Rep(H) → Rep(G). 2.5.9. Corollary. Assume that ϕ in 2.5.7 is surjective. Then (a) Pullback r 7→ ϕ∗ r establishes a bijection between representations of H and representations of G whose kernel contains ker ϕ. (b) This bijection respects irreducibility: Pullback restricts to a bijection between irreducible representations of H and irreducible representations of G whose kernel contains ker ϕ. (c) The inclusion of kernels ker ϕ ⊂ ker ρ holds if and only if ρ factors through ϕ. ∼

2.5.10. Remark. Let N := ker ϕ. When ϕ is surjective, one has G/N → H by the isomorphism theorem. Hence we can rephrase Corollary 2.5.9 in terms of (G, N ) as saying that representations of G/N correspond via pullback precisely to representations of G whose kernel contains N , and this correspondence respects irreducibility. 2.5.11. Proof of 2.5.7 and 2.5.9. For a subspace Z ⊂ W , stability under r(h) for all h ∈ H is at least as strong as stability under those r(h) of the form ϕ∗ r(g) = r(ϕ(g)) with g ∈ G. If r is surjective then every r(h) has the form r(ϕ(g)) for some g ∈ G. This proves 2.5.7(a) and 2.5.9(a)-(b). The factorization claims 2.5.7(b) and 2.5.9(c) follow from the more general statements for an arbitrary homomorphism ψ : G → K in place of the representation ρ. Certainly if ψ = f ◦ ϕ, then ker ϕ ⊂ ker ψ, since ϕ(g) = 1 implies ψ(g) = f (ϕ(g)) = f (1) = 1. If ϕ is surjective and the inclusion of kernels holds, then define f : H → K by f (h) = ψ(g) where g is arbitrary preimage of h in G under ϕ; this is well-defined because if g 0 is another such preimage then g −1 g 0 ∈ ker ϕ ⊂ ker ψ, hence ψ(g) = ψ(g 0 ). The diagram in the factorization question is (2.5.12)

 ψ

G ϕ

/K > ∃f ?

H 2.6. Applications of the 2nd orthogonality relation II: Comparing centralizers in G and its quotients. 2.6.1. Proposition (Comparing centralizers). Let G be a finite group and let N be a normal subgroup of G. For all g ∈ G, the centralizer of g in G is at least as large as the centralizer of gN in G/N . That is: | CentG (g)| ≥ | CentG/N (gN )|. 12

2.6.2. Proof. By the 2nd orthogonality relation 2.3.2, [g, g]Irr(G) = | CentG (g)| and every term χ(x)χ(x) in the sum defining [x, x] is ≥ 0. Similarly, working in G/N one has [gN, gN ]Irr(G/N ) = CentG/N (gN ). Since the irreducible characters of G/N are those of G whose kernel contains N (see 2.5.10), we conclude that [g, g]Irr(G) = [gN, gN ]Irr(G/N ) + ( terms each ≥ 0). So | CentG (g)| = [g, g]Irr(G) ≥ [gN, gN ]Irr(G/N ) = | CentG/N (gN )|.  2.6.3. Remark. Just before the result [9, 2.24], Isaacs remarks that it is possible but more tricky to prove the inequality of centralizers 2.6.1 directly without using characters. Can you see how to do it? 2.7. Tensor, symmetric and exterior powers: Characters. Compare with [15, 1.5-1.6]. 2.7.1. Review of the tensor product representation. Recall that if V, W are two representations of G, then the tensor product V ⊗ W is again a representation of G, where the action of g ∈ G on a simple tensor v ⊗ w is given by g(v ⊗ w) := gv ⊗ gw. Recall further that if (ei ) is a basis of V and (fj ) is a basis of W , then (ei ⊗ fj ) is a basis of V ⊗ W . 2.7.2. Lemma. The character of the tensor product is the product of the characters: χV ⊗W = χV χW 2.7.3. Motivation: Local vs. global again. The value of a character at each group element can be computed "locally" i.e., we can conjugate our linear transformation individually to make the computation more pleasant – we don’t need to worry what happens "globally" to the other elements, there is no problem of "simultaneous conjugation" here. 2.7.4. Proof. To compute χV ⊗W (g), choose bases (ei ) and (fj ) such that the action of g on V (resp. W ) in the basis (ei ) (resp. (fj )) is diagonal, i.e., we choose (ei ) and (fj ) to be bases of eigenvectors. That is, gei = λi ei and gfj = µj fj , where the λi (resp. µj ) are the eigenvalues of g acting on V (resp. W ). Then (ei ⊗ fj ) is a basis of eigenvectors for the action of g on V ⊗ W , with g(ei ⊗ fj ) = λi µj ei ⊗ fj . Therefore  ! X X X (2.7.5) χ(g) = λi µj = λi  µj  = χV (g)χW (g). i,j

i

j

 2.7.6. Symmetric and exterior powers. Compare [2, 11.5]. Let T n V := V ⊗n = V ⊗ · · · ⊗ V . | {z } n copies

Define the nth symmetric power Symn V as the quotient (2.7.7)

Symn V := T n V /span{v1 ⊗ · · · ⊗ vn − vσ(1) ⊗ · · · ⊗ vσ(n) | vi ∈ V, σ ∈ Sn }.

Define the nth exterior power ∧n V as the quotient (2.7.8)

∧n V := T n V /span{v1 ⊗ · · · ⊗ vn | vi ∈ V, vi = vj for some i 6= j}.

Then both Symn V and ∧n V are again representations of G, because in each case the subspace by which we have taken the quotient is G-stable.  2.7.9. The symmetric and exterior powers are defined as quotients. In characteristic 0, or more generally in characteristic p > n (so that (p, n!) = 1), the symmetric and exterior powers are also isomorphic (as representations of G) to subspaces of T n V . However, a symmetric power Symn V need not be isomorphic to a subrepresentation of T n V when p|n!. 13

2.7.10. Bases. Let (ei ) be a basis of V . Then the image of (ei1 ⊗ · · · ⊗ ein | i1 ≤ · · · ≤ id ) in Symn V , sometimes denoted ei1 · · · · · ein , gives a a basis of Symn V ; in particular   dim V + n − 1 n dim Sym V = . n Similarly, the image of (ei1 ⊗ · · · ⊗ ein | i1 < · · · < id ) n

in ∧ V , always denoted ei1 ∧ · · · ∧ ein , gives a basis of ∧n V ; in particular   dim V dim ∧n V = n and ∧n V = (0) for all n > dim V . 2.7.11. Lemma (Characters of the symmetric and exterior squares). One has (a) χSym2 V (g) =

χ(g)2 + χ(g 2 ) . 2

(b) χ∧2 V (g) =

χ(g)2 − χ(g 2 ) . 2

2.7.12. Proof. Set d = dim V . If (λ1 , . . . , λd ) are the eigenvalues of g acting on V (with each λi appearing as many times as the dimension of its eigenspace), then in view of the bases just described, the eigenvalues of g acting on Symn V are (λi1 · · · λin |i1 ≤ · · · ≤ in ),

(2.7.13)

i.e., the products of n eigenvalues, allowing repetitions. In particular, the eigenvalues of Sym2 V are the λi λj with i ≤ j. Hence X χSym2 V (g) = λi λj . i≤j

On the other hand, (2.7.14)

χ(g)2 =

X

λi

2

and χ(g 2 ) =

X

λ2i ,

since the eigenvalues of g 2 are the squares of the eigenvalues of g. Comparing gives (a). Similarly, the eigenvalues of g acting on ∧n V are (2.7.15)

(λi1 · · · λin | i1 < · · · < in ),

i.e., the products of n eigenvalues, not allowing repetitions (i.e., λ1 cannot be used twice, but it could be that λ1 = λ2 and then λ21 will appear) . Hence X χ∧2 V (g) = λi λj . i 1). By the Cauchy-Schwarz 17

Table 2.9.4. Character table of S5 completed: Decomposition of Sym2 Std (1)

(2)

(3)

(4)

(5)

(2)

(6)

S5

1(1)

(12)(10)

(123)(20)

(1234)(30)

(12345)(24)

(12)(34)(15)

(123)(45)(20)

1

1

1

1

1

1

1

1

sgn

1

−1

1

−1

1

1

−1

Std

4

2

1

0

−1

0

−1

sgn.Std

4

−2

1

0

−1

0

1

∧2 Std

6

0

0

0

1

−2

0

χ5 := Sym2 Std − 1 − Std

5

1

−1

−1

0

1

1

sgnχ5

5

−1

−1

1

0

1

−1

Sym2 Std

10

4

1

0

0

2

1

inequality, |ζ1 + · · · + ζd | ≤ |ζ1 | + · · · + |ζd | = d and there is equality in Cauchy-Schwarz if and only if the vectors are all in the same direction i.e., positive scalar multiples of one another (here we view the ζi as vectors in R2 ). Hence |χ(g)| = |χ(1)| = χ(1) implies that ζ1 = · · · = ζd . But then ρ(g) = ζ1 I and χ(g) = dζ1 . So χ(g) = χ(1) if and only if ζ1 = · · · = ζd = 1 i.e., ρ(g) = I.  We saw in the proof 3.1.4 that a special role is played not only by those g with χ(g) = χ(1), but also when merely the absolute values are equal, so we single this out: 3.1.5. Definition. The centralizer Z(χ) of a character χ is Z(χ) = {g ∈ G | |χ(g)| = χ(1)}.  As far as I know, the terminology "centralizer" is nonstandard for Z(χ); I just wanted to have some nonsymbolic way to call it. We will see that Z(χ) is related to the center Z(G) of G, so the terminology seems pertinent. For a group H, recall that X ∗ (H) is the group of linear characters of H. 3.1.6. Lemma. The centralizer Z(χ) of a character χ has the following properties: (a) Assume χ is afforded by a representation ρ : G → GL(V ). Then Z(χ) = {g ∈ G | ρ(g) scalar }. (b) The centralizer Z(χ) is a normal subgroup of G. (c) Define λ ∈ X ∗ (Z(χ)) by λ(g) = ζ if ρ(g) = ζI. Then ResG Z(χ) χ = χ(1)λ. (d) The quotient Z(χ)/ ker χ is cyclic. (e) One has Z(χ)/ ker χ ⊂ Z(G/ ker χ) (f ) If χ is irreducible, then equality holds in (e). 3.1.7. Remark. Recall that Z(GL(V )) consists precisely of the scalar matrices, so "ρ(g) scalar" is equivalent to "ρ(g) ∈ Z(GL(V ))". 3.1.8. Proof of 3.1.6. Part (a) was already proved in 3.1.4. Then (b)-(c) are clear. One has ker χ = ker λ; hence Z(χ)/ ker λ = Z(χ)/ ker χ. But Z(χ)/ ker λ ∼ = Im(λ), a finite subgroup of C× . Hence (d) follows from the fact that a finite subgroup of the multiplicative group of a field is cyclic. ∼ The isomorphism G/ ker χ → ρ(G) maps the subgroup Z(χ)/ ker χ onto ρ(Z(χ)). By (a), ρ(Z(χ)) = Z(GL(V )) ∩ ρ(G). Now Z(GL(V )) ⊂ Z(ρ(G)) gives (e). If χ is irreducible, then Z(ρ(G)) consists of scalar matrices by Schur’s Lemma. So we get the reverse inclusion Z(ρ(G)) ⊂ Z(GL(V )) ∩ ρ(G); whence (f).  18

3.2. Locating normal subgroups via the character table. 3.2.1. Lemma. Let ψ be a character of G. Then \

ker ψ =

ker χ

{χ∈Irr(G) | nχ (ψ)≥1}

3.2.2. Proof. The decomposition of ψ into irreducible characters is X ψ= nχ (ψ)χ. χ∈Irr(G)

If χ(g) = χ(1) for all χ with nχ (ψ) ≥ 1, the summing gives ψ(g) = ψ(1). To see the converse, observe the three relations: (a) |χ(g)| ≤ χ(1) P for all χ, (b) |ψ(g)| ≤P nχ (ψ)|χ(g)|, and (c) ψ(1) = nχ (ψ)χ(1). So to have the equality ψ(g) = ψ(1) one must have χ(g) = χ(1) for every χP with nχ (ψ) ≥ 1. (It is not necessary but one can also break the argument in two, which amounts to summing nχ (ψ)χ(g) in two steps – first the ‘inner sums’ which define the χ and then the sum over χ – by first deducing that |χ(g)| = χ(1) for all χ with nχ (ψ) ≥ 1.)  3.2.3. Question. Does the analogue of 3.2.1 hold for the centralizer Z(χ)?  3.2.4. No, the analogue can fail for the centralizers: We could have g ∈ G such that ρ1 (g) = ζ1 I and ρ2 (g) = ζ2 I with ζ1 = 6 ζ2 . Then g ∈ Z(χ1 ) ∩ Z(χ2 ) but g 6∈ Z(χ1 + χ2 ). 3.2.5. Corollary. The intersection of all the irreducible character kernels is trivial: \ ker χ = {1} χ∈Irr(G)

3.2.6. Proof. Revealed after asking the class. Apply 3.2.1 with ψ = χreg the regular character. The result follows because the regular representation is faithful and nχ (χreg ) = χ(1) for all irreducible χ.  3.2.7. Remark. For χ ∈ Irr(G), we find ker χ in the character table of G simply by locating the χ(1)’s in the row of χ and taking the union of those conjugacy classes. Recall that a subgroup N of G is normal if and only if N is a (disjoint) union of conjugacy classes of G. 3.2.8. Lemma. Every normal subgroup of G is an intersection of ker χ (over some subset of Irr(G)). More precisely: If N is normal in G with projection πN : G → G/N and regular character χreg,N , then \ N= ker χ. ∗ χ {χ∈Irr(G) | nχ (πN reg,N )≥1 }

∗ 3.2.9. Proof. Since the regular character χreg,N is faithful, its pullback πG χreg,N has kernel precisely N (recall 2.5). ∗ So the lemma follows from 3.2.1 applied with ψ = πG χreg,N .  ∗ Since pullback 2.5.9 identifies the χ satisfying nχ (πN χreg,N ) ≥ 1 with the χ satisfying N ⊂ ker χ, we can rephrase 3.2.8 as:

3.2.10. Corollary. N=

\

ker χ.

{χ∈Irr(G)|N ⊂ker χ}

For N = G0 the commutator subgroup (§1.7.3), the χ with G0 ⊂ ker χ have abelian image, hence are precisely the linear characters. Thus: 3.2.11. Corollary. The commutator subgroup of G is the intersection of all the degree 1 kernels: \ G0 = ker λ λ∈X ∗ (G)

Recall that a subgroup of G is characteristic if it is stable under all automorphisms of G. For example, the commutator subgroup G0 and the center Z(G) are characteristic in G. 3.2.12. Question. Can we tell from the character table which normal subgroups are characteristic? I don’t know. 19

3.3. Twisting representations by outer automorphisms. 3.3.1. Twisting on the source: twisting in G. If ρ : G → GL(V ) is a representation and ϕ ∈ Aut(G), then we can pullback ρ via ϕ to get a new representation ϕ∗ ρ := ρ ◦ ϕ of G; we also call ϕ∗ ρ the twist of ρ by ϕ. If ϕ ∈ Int(G) is inner, say ϕ = int(g), then intρ(g) defines an isomorphism between ρ and ϕ∗ ρ. But if ϕ is not inner (by slight abuse one often says ϕ is outer), then ϕ∗ ρ may very well be non-isomorphic to ρ. 3.3.2. Twisting in GL(V ). If ψ ∈ Aut(GL(V )), then we can instead form the pushout ψ∗ ρ := ψ ◦ ρ which is again a representation of G. Again if ψ is inner then it defines an isomorphism of ρ onto ψ∗ ρ, but if ψ is outer it might be that ψ∗ ρ ∼ 6 ψ. For example ψ="inverse transpose" is an outer automorphism of GL(n) for all n ≥ 2. Then ψ∗ ρ is = the dual of ρ; some representations are self-dual (isomorphic to their dual) while others are not. 3.3.3. Example: Z/p. Let G = Z/p be cyclic of prime order p. Then Aut(G) = (Z/p)× ∼ = Z/(p − 1). Further Aut(G) acts transitively on Z/p \ {0} by multiplication a · b = ab, so every non-trivial linear character of Z/p is a twist of every other by an automorphism: If ζ ∈ µprim and χa (1) = ζ a , then χb = ((ba−1 )∗ χa since p −1

((ba−1 )∗ χa (1) = χa (ba−1 ) = (ζ a )ba

= ζ b = χb (1).

4. April 9, 2020 4.1. Twisting representations by automorphism (continued). 4.1.1. Example: An . It is a basic theorem that every automorphism of Sn is inner unless n = 6 (one has Out(S6 ) = Z/2). Since Z(An ) = {1} for all n ≥ 4, one has Aut(An ) = Sn for all n ≥ 4. In particular, odd elements of Sn define outer automorphisms of Sn . Let σ ∈ Sn be odd. 4.1.2. Question. Given a representation ρ of An , when will (intSn σ)∗ ρ be isomorphic to ρ? More revealed after asking class. 4.2. The character table of A5 . We begin by restricting the irreducible representations of S5 to A5 . We find that Std and χ5 remain irreducible when restricted to A5 , while ∧2 Std has norm-squared 2, so ∧2 Std = χ3 + χ03 is a sum of two irreducible representations χ3 , χ03 . Checking that [∧2 Std, 1] = 0 shows that deg χ3 = deg χ03 = 3. 4.2.1. Using the orthogonality relations to find the remaining two rows. Applying the 2nd orthogonality relation gives 0 = [1, (123)] = 1 + 4 − 5 + 3χ3 ((123)) + 3χ03 ((123)). Hence χ3 ((123)) = −χ0 ((123)). Since the sum ∧2 Std((123)) = 0, we conclude that χ3 ((123)) = χ0 ((123)) = 0. Similarly, 0 = [(12)(34), 1] gives χ3 ((12)(34)) = χ03 ((12)(34)) = −1. Table 4.2.2. Character table of A5 Step 1: Restriction from S5 to A5 (1)

(3)

(5)

(5)

(2)

A5

1(1)

(123)(20)

(12345)(12)

(12354)(12)

(12)(34)(15)

ResSA55 sgn = ResSA55 1 = 1

1

1

1

1

1

ResSA55 Std = ResSA55 sgnStd

4

1

−1

−1

0

ResSA55 χ5 = ResSA55 sgnχ5

5

−1

0

0

1

factor of ResSA55 ∧2 Std

3

factor of ResSA55 ∧2 Std

3

ResSA55 ∧2 Std

6

0

1

1

−2

So far we have reached Table 4.2.3. To fill the last 4 entries, we use the 1st orthogonality relation. It is important to observe that 5-cycles are conjugate to their own inverses, even in A5 , so these 4 entries are real. Let χ3 ((12345)) = x and χ03 ((12345)) = y. Then 1 = [χ3 , χ3 ] gives x2 + y 2 = 3, 20

Table 4.2.3. Character table of A5 Step 2: Preparing for the final ascent (1)

(3)

(5)

(5)

(2)

A5

1(1)

(123)(20)

(12345)(12)

(12354)(12)

(12)(34)(15)

ResSA55 sgn = ResSA55 1 = 1

1

1

1

1

1

ResSA55 Std = ResSA55 sgnStd

4

1

−1

−1

0

ResSA55 χ5 = ResSA55 sgnχ5

5

−1

0

0

1

factor of ResSA55 ∧2 Std

3

0

x

y

−1

factor of ResSA55 ∧2 Std

3

0

ResSA55 ∧2 Std

6

0

−1 1

−2

1

while 0 = [χ3 , 1] gives x + y = 1. Hence x, y are the roots of t2 − t − 1 ∈ Z[t]. So

√ 1± 5 . 2

x, y =

4.2.4. Conclusions about A5 . We conclude that the character table of A5 is as given in Table 4.2.5. (a) As we suspected the two irreducible characters of degree 3 are twists of one another by an automorphism of A5 which is outer as automorphism of A5 but inner as automorphism of S5 , i.e., conjugation by an odd permutation. (b) If a conjugacy class K of An is also a conjugacy class of Sn , then a character χ and its twist int(σ)∗ χ by odd σ will have same value on K , because then int(σ)(K ) = K . Write χ3 for one of the irreducible characters of degree 3. What tells us that int((45))∗ χ3 6= χ3 is that they differ on the unique pair of conjugacy classes of A5 which are not conjugacy classes of S5 (but whose union is the class of 5-cycles in S5 ). (c) That one conjugacy class of S5 breaks up into two of A5 seems to correspond to one irreducible character of S5 breaking up into two upon restriction to A5 . This is an interesting example of the mysterious duality we see between conjugacy classes and irreducible characters. Table 4.2.5. Character table of A5 completed (1)

(3)

(5)

(5)

(2)

A5

1(1)

(123)(20)

(12345)(12)

(12354)(12)

(12)(34)(15)

ResSA55 sgn = ResSA55 1 = 1

1

1

1

1

1

ResSA55 Std = ResSA55 sgnStd

4

1

−1

−1

0

ResSA55 χ5 = ResSA55 sgnχ5

5

−1

0

0

1

χ3

3

0

√ 1+ 5 2

√ 1− 5 2

−1

√ 1+ 5 2

−1

1

−2

int((45))∗ χ3

3

0

√ 1− 5 2

ResSA55 ∧2 Std

6

0

1

21

4.2.6. Finding χ3 via eigenvalues. Instead of using the orthogonality relations, one can also determined the last two rows of the character table 4.2.5 of A5 by arguing in terms of the roots of unity which are the eigenvalues in a representation ρ3 : A5 → GL(3, C) affording χ3 . Let ζp denote a primitive pth root of unity. Then ζpj + ζp−j is real (for example if ζp = e2πi/p , then P ζpj + ζp−j = 2 cos(2jπi/p). Conversely, if a sum aj ζpj is real then one has aj = ap−j = a−j , where j is naturally considered mod p (i.e., in Fp ) since ζpp = 1. Using a bit more field theory language, one says that Q(ζp + ζp−1 ) is the maximal totally real subfield of Q(ζp ); the subfield Q(ζp + ζp−1 ) is the fixed field of the automorphism ζp 7→ ζp−1 ; if we think of Q(ζp ) as a subfield of C then this automorphism is complex conjugation. For each g ∈ A5 , the image ρ3 (g) has 3 eigenvalues counting multiplicity which are all |g|th roots of unity. We check that g is conjugate to g −1 for all g ∈ A5 , so χ(g) is real for every character χ of A5 . Finally χ3 (g) < 3 for all g ∈ A5 \ {1} because A5 is simple, so ker χ3 = {1}. Hence: (a) ρ3 ((123)) must have eigenvalues ζ3 , ζ32 , 1 (each with multiplicity 1). So χ3 ((123)) = ζ3 + ζ22 + 1 = 0 (b) The eigenvalues of ρ3 ((12)(34)) are ±1. Since A5 is simple the centralizer Z(χ3 ) is trivial, so we cannot have eigenvalue 1 or −1 with multiplicity 3. Hence we either have eigenvalue 1 or −1 with multiplicity 2 and the other with multiplicity 1. So χ3 ((12)(34)) ∈ {±1} (c) ρ3 ((12345)) must have eigenvalues ζ5 , ζ5−1 , 1 or ζ52 , ζ5−2 , 1. If ρ3 ((12345)) has eigenvalues ζ5 , ζ5−1 , 1, then ρ3 ((12345)) must have eigenvalues ζ52 , ζ5−2 , 1, else [χ3 , χ3 ] would not be integral. (d) We can now conclude from [χ3 , 1] = 1 that χ3 ((12)(34)) = −1 rather than 1. (e) Note that what we call χ3 and what we call int(45)∗ χ3 is just a question of labelling; these two characters are algebraically indistinguishable. 4.3. More about character centralizers. We do have an analogue of the intersection of all irreducible kernels being trivial 3.2.5 for centralizers: 4.3.1. Corollary. Z(G) =

\

Z(χ).

χ∈Irr(G)

4.3.2. Proof. Let χ be irreducible and let π : G → G/ ker χ be the projection. One has (H. ker χ)/ ker χ = π(H) for every subgroup H of G. Take H = Z(G). Since π is surjective, π(Z(G)) ⊂ Z(π(G)). That is, (Z(G). ker χ)/ ker χ ⊂ Z(G/ ker χ). Recall that Lemma 3.1.6(f) states that Z(G/ ker χ) = Z(χ)/ ker χ. Putting the last two together gives (Z(G). ker χ)/ ker χ ⊂ Z(χ)/ ker χ). By the isomorphism theorem, this entails Z(G). ker χ ⊂ Z(χ); in particular Z(G) ⊂ Z(χ). The converse is a bit more interesting. Assume g ∈ Z(χ) for all χ ∈ Irr(G). Then π(g) ∈ π(Z(χ)) ⊂ Z(π(G)) by Lemma 3.1.6(e). It follows that, for all x ∈ G, π[x, g] = [π(x), π(g)] = 0 in π(G) = G/ ker χ. That is, [x, g] ∈ ker χ for all χ ∈ Irr(G). But we have seen (3.2.5) that the intersection of all the irreducible kernels is trivial. Hence [x, g] = 1 and g commutes with all x ∈ G.  4.4. Existence of a faithful irreducible character. The attributes "faithful" and "irreducible’ are quite different. For example, the irreducible characters generate all the characters in the additive sense of forming a basis of the space of class functions, while a faithful representation ρ generates all representations in the multiplicative sense that any other representation is a constituent of some tensor power of ρ. The relationship between the two notions is not so simple. Nevertheless, under additional hypotheses, there are certain implications: 4.4.1. Theorem. Assume G admits a faithful irreducible representation. Then the center Z(G) is cyclic. 4.4.2. Proof 1. Assume ρ : G → GL(V ) is faithful and irreducible. Since ρ is faithful, it embeds Z(G) into GL(V ). Since Z(G) ⊂ GL(V ) commutes with the image of G, Schur’s Lemma implies that every element of Z(G) is scalar in GL(V ). Thus Z(G) ⊂ Z(GL(V )). But Z(GL(V )) ∼  = C× and every finite subgroup of C× is cyclic. The following proof, given in [9] illustrates the use of the centralizers Z(χ). 22

4.4.3. Proof 2. Let χ = χρ be the character of ρ. Since ρ is faithful, ker χ = {1} and 3.1.6(d),(f) simplify to Z(χ) cyclic and Z(G) = Z(χ) respectively.  In the positive direction, the following shows that, for a p-group, a cyclic center is also sufficient for the existence of a faithful irreducible representation. 4.4.4. Theorem. Assume G is a p-group with cyclic center. Then G admits a faithful irreducible representation. More precisely, if χ is an irreducible character and ker χ does not contain the unique order p subgroup of Z(G), then χ is faithful. First we recall a basic lemma about normal subgroups of p-groups: 4.4.5. Lemma. Let G be a p-group. Then every nontrivial normal subgroup of G intersects the center Z(G) nontrivially. 4.4.6. Proof. Suppose N is normal in G. Then N is a disjoint union of conjugacy classes Ki of G. Since |Ki | = |G|/| CentG (Ki )| (where | CentG (Ki )| is the common value of CentG (x) for all x ∈ Ki ), a conjugacy class Ki in a p-group is either a singleton of an element of the center, or its size is divisible by p. Since N is divisible by p and N contains at least one class of size one, namely the identity {1}, it must contain at least p classes of size one.  4.4.7. Proof of 4.4.4. Since Z(G) is a cyclic p-group, it contains a unique subgroup Z ⊂ Z(G) of order p. Let N be nontrivial and normal in G. By Lemma 4.4.5, the intersection N ∩ Z(G) is nontrivial. So it contains a subgroup of order p and the only option is Z. So every nontrivial normal subgroup contains N . Since the intersection of all the irreducible kernels is trivial (3.2.5) ker χ does not contain Z for some χ ∈ Irr(G). But ker χ is a normal subgroup, so it must be trivial.  4.4.8. Remark. If G is not a p-group, it can happen that Z(G) is cyclic and yet G does not have a faithful irreducible representation. See [9, Problem 2.19] for an example. 4.5. Bounding character degrees. A basic part of understanding the set Irr(G) of irreducible characters is to understand the set of degrees {χ(1) | χ ∈ Irr(G)}. Two of the simplest questions which come to mind are: 4.5.1. Question (Size). What can we say about the size of the χ(1) for irreducible χ? How large/small can they be? 4.5.2. Question (Divisibility). What are the divisibility properties of the irreducible character degrees? Regarding divisibility properties, as Ludvig pointed out in an early lecture, we will soon show (with the aid of algebraic integers): 4.5.3. Theorem. The degree of an irreducible character divides the order of the group: χ(1) | |G| for all χ ∈ Irr(G). The proof is given in 6.4.1. In fact, with more work one can do even better and show that χ(1) divides the index [G : Z(χ)], see 7.1.2. For now, let us prove the corresponding inequality: 4.5.4. Theorem. Let χ ∈ Irr(G). (a) One has χ(1)2 ≤ [G : Z(χ)]. (b) Equality holds if and only if χ(g) = 0 for all g ∈ G \ Z(χ). First we bound the norm-squared |χ|2 = [χ, χ] of a restriction in terms of the index. Intuitively, we expect that a restriction of an irreducible character is likely to become reducible, so expect the norm-squared of the restriction to grow – the smaller the subgroup the more reducible we expect the restriction to be, so the larger we expect the norm-squared to be. The following lemma says it doesn’t grow too much. 4.5.5. Lemma. Assume χ ∈ Irr(G) is an irreducible character of G and that H is a subgroup of G. Then (a) 2 |resG H χ| ≤ [G : H] (b) Equality holds in (a) if and only if χ(g) = 0 for all g ∈ G \ H. P P 2 2 2 2 4.5.6. Proof. By definition |H||ResG H χ| = h∈H |χ(h)| and |G||χ| = g∈G |χ(g)| = |G| since G is irreducible. Since |χ(g)|2 ≥ 0 for all g ∈ G \ Z(χ), we conclude that 2 |H||ResG H χ| ≤ |G|



with equality as described in (b). 23

4.5.7. Proof of 4.5.4. We apply the lemma with H = Z(χ). Recall from Lemma 3.1.6 that ResG Z(χ) = χ(1)λ where λ is a linear character of Z(χ). Since |λ|2 = 1, one has 2 χ(1)2 = χ(1)2 |λ|2 = |ResG Z(χ) | ≤ [G : Z(χ)]

and equality holds precisely when claimed. The following illustrates a group-theoretic situation where we do have equality in Theorem 4.5.4:



4.5.8. Theorem. Let χ ∈ Irr(G). Assume the quotient G/Z(χ) is abelian. Then χ(1)2 = [G : Z(χ)]. 4.6. More about permutation representations. We have already used repeatedly the permutation representation ρ0 of Sn acting (doubly-transitively, in fact n-transitively) on {1, 2, . . . , n} as a means of constructing an interesting irreducible representation, namely ρ such that ρ0 = ρ ⊕ 1. Recall that the characters χ of ρ and χ0 = Std of ρ0 satisfy, respectively, (4.6.1a)

χ0 (σ) = |Fix(σ)| where Fix(σ) = {a ∈ {1, 2 . . . , n} | σ(a) = a}, and

(4.6.1b)

χ(σ) = |Fix(σ)| − 1.

More generally, given an action of a finite G on a set X, we would we would like to understand the character of the corresponding linear representation of G. We consider the special case when X := Class(H) := {gHg −1 | g ∈ G} is the conjugacy class of some subgroup H of G. Recall that, if G acts on itself by left multiplication and on Class(H) by conjugaction, then G → Class(H) x 7→ xHx−1 is a G-equivariant map which induces an isomorphism of G-sets ∼

G/NG (H) → Class(H) x 7→ xHx−1 . The following analogue of Burnside’s Lemma allows us to compute the character of χ. 4.6.2. Lemma. Let G acts on Class(H) by conjugation. For every σ ∈ G, one has (4.6.3)

|Fix(σ)| =

|ClassG (σ) ∩ NG (H)| · |ClassG (H)| . ClassG (σ)

4.6.4. Proof. First recall that x 7→ gx is a bijection between Fix(σ) and Fix(gσg −1 ), while σ 7→ gσg −1 is an isomorphism between StabG (x) and StabG (gx) which restricts to a bijection between StabG (x) ∩ Class(σ) and StabG (gx) ∩ Class(σ). So |Fix(τ )| is independent of τ ∈ Class(σ) and | Stab(x) ∩ Class(σ)| is independent of x ∈ Class(H), both because the action of G on Class(H) is transitive. Now we count X |Fix(τ )| τ ∈Class(σ)

in two ways as in the proof of Burnside’s Lemma. On the one hand, summing |Fix(τ )| over τ gives |Class(σ)||Fix(σ)|. On the other hand, summing over Class(H), each element of Class(H) is stabilized by |Class(σ) ∩ NG (H)| many elements of Class(σ). Hence |Class(σ)||Fix(σ)| = Class(H)||NG (H) ∩ Class(H)| and solving for |Fix(|sigma)| gives the result. 4.6.5. The action of S5 on its 3-Sylows. 4.6.6. The action of S5 on its 5-Sylows. Revealed after Homework 4 is due. 4.6.7. The action of S7 on its 7-Sylows. Let P be a 7-Sylow in S7 . Then the normalizer N := NS7 (P ) is a Frobenius group of order 42. It has the form N = P.H, where H is cyclic of order 6 generated by a 6-cycle. So N contains 7 elements of cycle type (2, 2, 2), 14 elements of type (3, 3) and the number of 6-cycles in N is 7. Plugging this into the Burnside-type Lemma 4.6.2 allows to compute the permutation character χ(g) = |Fix(g)| − 1 for the action of G by conjugation on Class(P ). If I didn’t make a mistake, one finds [χ, χ] = 6. 24

4.7. Other problems from Chapter 2 of [9]. 4.7.1. Lemma ([9], (2.1)(a)). Let F be a field (not necessarily algebraically closed, not necessarily of characteristic zero). Assume ρ : G → GL(V ) is a nontrivial, irreducible representation of G over F . Then X (4.7.2) ρ(g) = 0. g∈G

4.7.3. Proof. Let α be the sum (4.7.2). As usual, we also write ρ : F [G] → End(V ) for the linear extension of P the group representation to a module of the group algebra F [G]. Since α = ρ( g∈G g), we see that α lies in the centralizer CentEnd(V ) ρ(F [G]) of the F -algebra ρ(F [G]). For all h ∈ G, one has X X ρ(h)α = ρ(hg) = ρ(t) = α, g∈G

t∈G

since g 7→ hg is a bijection G → G. Hence α(ρ(h) − I) = 0. ⊕ dim V ∼ . Since ρ is irreducible but not the trivial representation, there exists If ρ(h) = I for all h ∈ G, then ρ = 1 h ∈ G such that ρ(h) 6= I. Since ρ is irreducible, Schur’s Lemma implies that the centralizer CentEnd(V ) ρ(F [G]) is a division ring. Thus, if α 6= 0, then α is invertible in CentEnd(V ) ρ(F [G]) and we get ρ(h) − I = 0, contradiction. Hence α = 0.  4.7.4. Remark. It could be useful to see how the argument becomes conceptually simpler if we assume in addition that F is algebraically closed. Then Schur’s Lemma implies that α is scalar, and one sees more directly that α is either invertible or zero. 4.7.5. Corollary ([9] (2.1)(b)). Assume ρ : G → GL(V ) is a semisimple representation which does not contain 1. Then (4.7.2) holds. L ⊕ni 4.7.6. Proof. Write ρ = ρi as a direct sum of irreducible representations ρi . By assumption ρi ∼ 6 1 for all i. So = L ⊕ni if α(ρ) denotes the sum (4.7.2), then α(ρi ) = 0 for all i by Lemma 4.7.1. But α(ρ) = α(ρi ) so α(ρ) = 0.  We deduce the following criterion to show that a character vanishes at a given element: 4.7.7. Corollary ([9], (2.1)(b)). Let χ be a complex character of G. Assume there exists a subgroup H of G an element g ∈ G satisfying (a) All elements of the coset gH are conjugate in G, and (b) [ResG H χ, 1] = 0. Then χ(g) = 0. 4.7.8. Proof. By hypothesis (b), Corollary 4.7.5 applies to the H-character ResG H χ. Let ρ afford χ. Then X α(ResG ρ(h) = 0. H ρ) = h∈H G So on the one hand ρ(g)α(ResG H ρ) = 0 and thus tr[ρ(g)α(ResH ρ)] = 0. But on the other hand X X tr[ρ(g)α(ResG ρ(gh)] = χ(gh), H ρ)] = tr[ h∈H

and hypothesis (a) implies that

P

h∈H

h∈H

χ(gh) = |H|χ(g). Putting the two together gives χ(g) = 0.

4.8. Applications of the 2nd orthogonality relation III : Real elements. 4.8.1. Definition. An x ∈ G is called a real element of G if x is conjugate to its inverse x−1 . 4.8.2. Proposition (Problem (2.11) in Isaacs [9]). An element x ∈ G is real if and only if χ(x) ∈ R is real for all χ ∈ Irr(G). 4.8.3. Proof. Revealed after the homework was due. If x is real then χ(x) = χ(x−1 ) = χ(x) is real for every character χ. Conversely, assume that χ(x) ∈ R for all χ ∈ Irr(G). Then χ(x)χ(x−1 ) = χ(x)χ(x) = χ(x)2 is real and non-negative (as the square of a real number). Hence [x, x−1 ] is a sum of non-negative real terms χ(x)2 and one of those terms is 1(x)2 = 1 > 0. Hence [x, x−1 ] > 0, so x is conjugate to x−1 by the 2nd orthogonality relation 2.3.2.  25

5. April 16, 2020 5.1. Explicit tensor-generation: The Brauer-Burnside theorem. 5.1.1. Theorem. Let χ be a faithful character of G. Assume χ : G → C takes on precisely m values. Then every irreducible ψ ∈ Irr(G) is a constituent of some χj for some 0 ≤ j ≤ m − 1. For a related example, see Problem 5 on Homework 4. 5.1.2. The Vandermonde determinant. Given a commutative ring R and a sequence a = (a1 , . . . , an ) ∈ Rn , the Vandermonde matrix of a is   1 a1 a21 · · · an−1 1  1 a2 a22 · · · an−1  2   (5.1.3) V (a1 , . . . , an ) =  . .. .. ..  ..  .. . . . .  2 n−1 1 an an · · · an and the Vandermonde determinant V (a1 , . . . , an ) := det V (a1 , . . . , an ) is the determinant of the Vandermonde matrix. 5.1.4. Lemma (Vandermonde’s identity). Assume R is an integral domain. For all n ≥ 2, one has n(n−1) Y V (a1 , . . . , an ) = (−1) 2 (ai − aj ). i 1 is then called a Salem number. The most famous example of Salem number is the Lehmer’s number; it is the largest real root of Lehmer’s polynomial x10 + x9 − x7 − x6 − x5 − x4 + x + 1. It is a longstanding open conjecture that Lehmer’s number is the smallest Salem number. For those interested in this direction, see also "Mahler measure". 6.3. Some of Burnside’s applications of integrality. 6.3.1. Theorem (Burnside). Let χ ∈ Irr(G) and K ∈ Class(G) a conjugacy class of G. Assume (χ(1), |K |) = 1. Then either χ(K ) = 0 or K ⊂ Z(χ). Note that Z(χ) is normal in G, so a conjugacy class is either contained in it, or the two have empty intersection. 6.3.2. Proof. By the central character integrality Theorem 6.1.4, χ(K )|K | ∈Z χ(1) is an algebraic integer. By the generalized Euclid lemma 6.2.4, the assumption (χ(1), |K |) = 1 implies that also χ(K )/χ(1) ∈ Z is integral. If K ⊂ Z(χ), we are done; so assume K ∩ Z(χ) = ∅. Put α = χ(K )/χ(1). Then α is integral and |α| < 1, so the "roots of unity averages" lemma 6.2.6 applies to it; hence α = 0. So χ(K ) = 0.  6.3.3. Theorem. If G is nonabelian simple, then the only conjugacy class in G of prime power order is the trivial one {1}. 33

6.3.4. Proof. Assume p is prime and K is a conjugacy class of order pj for P some j > 1. Since K 6= {1} the regular character satisfies χreg (K ) = 0. Since χreg = χ∈Irr(G) χ(1)χ, evaluation at K gives X (6.3.5) χ(1)χ(K ) = 0 χ∈Irr(G)

Since G is nonabelian simple, Z(χ) = Z(G) = {1} for every nontrivial irreducible character χ. If (p, χ(1)) = 1 for an irreducible χ, then Theorem 6.3.1 implies χ(K ) = 0. So the terms in (6.3.5) with p not dividing χ(1) and χ nontrivial are zero. Separating the trivial character from the rest gives X χ(1)χ(K ) = −1. χ∈Irr(G), p|χ(1)

Hence X χ∈Irr(G), p|χ(1)

χ(1) 1 χ(K ) = − . p p

The left-hand side is an algebraic integer while the right-hand side is not; contradiction.



6.3.6. Theorem (‘Burnside’s pa q b Theorem’). Assume G is a finite group whose order is pa q b for some primes p, q and positive integers a, b. Then G is solvable. 6.3.7. Proof. Recall that, given a normal subgroup N of G, one has that G is solvable if and only if both N and 0 0 the quotient G/N are solvable. Since every subgroup and quotient of G also has order of the form pa q b , we see by induction that it is equivalent to show that G is not nonabelian simple. Assume G is simple. We show G is abelian (hence cyclic of prime order). Let P ∈ Sylp (G) be a p-Sylow subgroup. Recall that p-groups have nontrivial center; choose z ∈ Z(P ) \ {1}. Then the centralizer CentG (z) contains P , so |Class(z)| = |G|/| CentG (z) divides q b ; in particular the conjugacy class of z has prime power order. Since z 6= 1, Theorem 6.3.3 implies G is abelian.  6.3.8. Remark. We know that the smallest nonabelian finite simple group is A5 ; its order 60 = 22 · 3 · 5 is divisible by 3 primes. The next two finite nonabelian simple groups are GL(F2 ) ∼ = PSL(2, F7 ) and A6 ∼ = PSL(2, F9 ) of orders 3 3 2 168 = 2 · 3 · 7 and 360 = 2 · 3 · 5 respectively, also only divisible by 3 primes. I think there are only finitely many finite simple groups whose order is divisible by precisely three primes and perhaps the above 3 are all of them(?). Discuss |An |, |PSL(2, Fpn )|. 6.4. More integrality: Irreducible character degrees divide the order of the group. 6.4.1. Proof of 4.5.3. By now this is mostly a matter of recollections: First, recall the statement: If χ ∈ Irr(G), then the degree χ(1) divides the order |G|. Second, recall that, by the 1st orthogonality relation stated in terms of conjugacy classes 1.4.10, the irreducibility of χ implies s X |G| = [χ, χ]|G| = |Kj | χ(Kj )χ(Kj−1 ). j=1

Third, recall the relation (6.1.3) between the central character on the class sums Kj and the character values: |Kj |χ(Kj ) = χ(1)ωχ (Kj ) Using this relation to rephrase the orthogonality as s X |G| = χ(1)ωχ (Kj )χ(Kj−1 ). j=1

Dividing by χ(1) gives s

X |G| = ωχ (Kj )χ(Kj−1 ). χ(1) j=1 The right-hand side is an algebraic integer by Theorem 6.1.4. Hence |G|/χ(1) ∈ Z ∩ Q = Z.



6.4.2. Remark. It’s interesting that the proof uses algebraic integers even though the statement 4.5.3 is purely in terms of ordinary integers. As we have previously remarked, one can work harder and show that χ(1) divides the index [G : Z(χ)] but we probably won’t prove that (see [9, (3.12)]). Let us however deduce the following interesting group-theoretic consequence from the strong statement χ(1)|[G : Z(χ)]: 34

6.4.3. Theorem. Let p be prime. Assume G has (a) a faithful irreducible character χ of degree pa for some a ≥ 1, and (b) an abelian p-Sylow subgroup. Then pa is the exact power of p dividing |G/Z(G)|. 6.4.4. Proof. Since χ is faithful Z(χ) = Z(G) (Lemma 3.1.6 on character centralizers). By the strengthening of Theorem 4.5.3 which we are assuming, χ(1) = pa divides |G/Z(G)|. We are left to show that pa+1 doesn’t divide |G/Z(G)|. Let P Sylp (G) and x ∈ P . Since P is assumed abelian, CentG (x) contains P . So the hypothesis (χ(1), |class(x)|) = 1 of the relatively prime theorem 6.3.1 is satisfied. Hence χ(x) = 0 for all x ∈ G \ Z(G).  6.5. Applications of the 2nd orthogonality relation IV: Real-valued characters and odd-order groups. Recall that the very difficult ‘Odd order theorem’ of Feit-Thompson asserts that every finite group of odd order is solvable. In the expository paper [4], Glauberman explains some of the history and key ideas of the odd order theorem, by describing some much easier results which use odd order in an essential way. The following is one half of [4, Th. 2]. We explain the proof given in Glauberman’s reference, namely [7, Th. 3.6], since it is again a nice application of the second orthogonality relation. It also illustrates the technique of playing off (i) that character values are algebraic integers versus (ii) that an algebraic integer which is a rational number is an integer i.e., lies in Z. Moreover, the use of this technique is more elementary here than in the proof of Burnside’s pa q b theorem (see 6.3.6). 6.5.1. Theorem (Burnside, [7], Th. 3.6). Assume G has odd order. Then: (a) The only real element of G is the identity (compare 4.8.1). (b) The only real-valued irreducible character of G is the trivial character. 6.5.2. Proof of (a). Recall that, in a cyclic group C of order n, generated by x ∈ C, the generators of C are precisely the powers xj with j relatively prime to n. In particular, if n is odd, then x2 is a generator, so x is a power of x2 (specifically x = (x2 )(n+1)/2 ). Assume g is conjugate to g −1 , say int(x)g = g −1 . Since int : G → Int(G) is a homomorphism, x2 commutes with g. But then so do all its powers, which include x since G has odd order. Hence int(x)g = g, so g 2 = 1. Since G has odd order, g = 1.  6.5.3. Proof of (b). Assume χ is a nontrivial, real-valued irreducible character. By Theorem 4.5.3, the degree χ(1) is odd. Since χ 6= 1, the 1st orthogonality relation gives X 0 = [χ, 1] = χ(g). g∈G

We break the sum into χ(1) and the rest of the terms. This gives X (6.5.4) χ(g) = −χ(1). g∈G\{1} −1

Since χ is real, χ(g) = χ(g ) for all g ∈ G. Since G is of odd order, g 6= g −1 for all non-identity g ∈ G. So we can partition G \ {1} as G \ {1} = S t S −1 where each pair {g, g −1 } intersects both S and S −1 in precisely one element. Then (6.5.4) gives X χ(1) . χ(g) = − 2 g∈S

As a sum of character values, the left-hand side is an algebraic integer. By contrast, since G has odd order, the right-hand side is a rational number which is not an integer; contradiction. 7. April 30, 2020 7.1. Reduction of χ(1) divides [G : Z(χ)] to χ(1) divides |G| via tensor products (d’après Tate). Let χ ∈ Irr(G) be an irreducible character. There is an argument in Serre [15, §6.5, Prop. 17], which Serre attributes to Tate, which uses tensor products to reduce the stronger divisibility relation χ(1)|[G : Z(G)] to the one have already proved, that the degree divides the order of the group (4.5.3, 6.4.1). Serre does not discuss the centralizers Z(χ) in his book, but we can further reduce divisibility of [G : Z(χ)] to that of [G : Z(G)]. 7.1.1. Remark. John Tate was one of the giants of twentieth century number theory and algebraic geometry. Sadly he passed away in October 2019. 7.1.2. Theorem. Assume χ ∈ Irr(G). Then χ(1) divides the index [G : Z(χ)]. 35

7.1.3. Reduction to χ(1) divides [G : Z(G)]. Assume χ(1) divides |G/Z(G)| for all (G, χ); we deduce the same with G/Z(G) replaced by G/Z(χ). Any representation ρ : G → GL(V ) affording χ factors through ρ : G/ ker χ = G/ ker ρ : G → GL(V ). That is, it is in the image of the pullback map 2.5.9. The character χ of ρ is still irreducible; what we have gained is that we know it’s faithful. We deduce from our assumption that χ(1) = χ(1) divides [G/ ker χ : Z(G/ ker χ)].

(7.1.4)

Since χ ∈ Irr(G/ ker χ) is faithful and irreducible, the properties of the centralizer Z(χ) (Lemma 3.1.6) imply that Z(χ) = Z(G/ ker χ) = Z(χ)/ ker χ. Plugging back into (7.1.4) gives χ(1) divides [G/ ker χ : Z(χ)/ ker χ]. Finally, simplifying fractions yields: [G/ ker χ : Z(χ)/ ker χ] = [G : Z(χ)].  7.1.5. Small upgrade of tensor products. Given ρ : G → GL(V ) and σ : G → GL(W ) we have previously viewed the tensor product as ρ ⊗ σ : G → GL(V ⊗ W ) where the action on a simple tensor v ⊗ w is given by ρ ⊗ σ(g)(v ⊗ w) = ρ(g)v ⊗ σ(g)w. For the next argument, we observe that we can also view ρ ⊗ σ : G × G → GL(V ⊗ W ) where now (g, h) ∈ G × G acts on v ⊗ w by ρ(g)v ⊗ σ(h)w. To see that two actions yield an action of the direct product amounts to checking that the actions commute, i.e., that the relations (g, 1)(1, h) = (1, h)(g, 1) = (g, h) continue to hold on the target GL(V ⊗ W ). This is clear in our context because the different factors of G are acting on different factors of the tensor product. It is similar to the commutation of the left and right multiplication actions of G which was used in studying the basic ring theory of C[G]. In terms of the upgraded G × G → GL(V ⊗ W ), the previous G → GL(V ⊗ W ) is the composition G

∆

/ G×G

ρ⊗σ

/ GL(V ⊗ W )

where ∆ : G → G × G is the diagonal embedding g 7→ (g, g). For the same reason as before (2.7.2), the character of ρ ⊗ σ : G × G → GL(V ⊗ W ) is χρ χσ : G × G → (g, h) 7→

C χρ (g)χσ (h)

7.1.6. Lemma. A tensor product ρ⊗σ : G×G → GL(V ⊗W ) is irreducible if and only if both factors ρ : G → GL(V ) and σ : G → GL(W ) are irreducible. 7.1.7. Proof. Here is a proof using characters and the orthogonality relations. The conjugacy classes of G × G are precisely the X × Y , where X, Y is a pair of (not necessarily distinct) conjugacy classes in G. Hence the number of classes in G × G is s2 , where we denote the number of classes in G by s as before. Given characters χ, ψ of G, the character χψ : G × G → C satisfies   ! X X X 1 1 1 [χψ, χψ]G×G = |χ(g)ψ(h)|2 =  |χ(g)|2  |ψ(h)|2 = [χ, χ]G · [ψ, ψ]G . |G|2 |G| |G| (g,h)∈G×G

g∈G

h∈G

Therefore χψ is irreducible as character of G × G if and only if χ, ψ are both irreducible characters of G. So the χψ give s2 irreducible characters of G × G, and they are all there is because the number of irreducible characters equals the number of conjugacy classes.  36

7.1.8. Tate’s proof that χ(1) divides |G/Z(G)|. Assume ρ : G → GL(V ) affords χ ∈ Irr(G). For every n ≥ 1, consider ρ⊗n : Gn → GL(V ⊗n ), where Gn := G × · · · × G and V ⊗n := V ⊗ · · · ⊗ V {z } {z } | | n copies

n copies

One has Z(Gn ) = Z(G)n . Let Z(G)n0 := {(z1 , . . . zn ) ∈ Z(G)n |

n Y

zi = 1};

i=1

this is a multiplicative version of the "sum of coordinates equal zero" vector subspace we considered long Qn ago. In particular, the order of Z(G)n0 is |Z(G)|n−1 , because the map Z(G)n → Z(G) given by (z1 , . . . , zn ) 7→ i=1 zi is a surjective homomorphism with kernel Z(G)n0 . Since χ is irreducible, Schur’s Lemma implies that every z0 ∈ Z(G) acts by a scalar – the value ωχ (z0 ) of the central of the tensor product, it follows that z = (z1 , . . . , zn ) ∈ Z(G)n acts on V ⊗n Qn character at z0 . By definition n ⊗n by i=1 ωχ (zi ). Hence Z(G)0 ⊂ ker ρ acts trivially. So ρ⊗n factors as ρ⊗n

Gn $

n

G

/ GL(V ⊗n ) . 8 ρ⊗n

/Z(G)n0

Since ρ⊗n , and hence also ρ⊗n , is irreducible (7.1.6), we may apply "the degree divides the order" (4.5.3) to ρ⊗n ; we get |G|n (dim V )n divides . |Z(G)|n−1 Hence  n |G|n |G| = |Z(G)| ∈ Z |Z(G)|n−1 (dim V )n |Z(G)|(dim V ) or in other words n  1 |G| ∈ Z |Z(G)|(dim V ) |Z(G)| for all n ≥ 1. It follows from the "finitely generated module" characterization of algebraic integers (5.4.7 that   |G| ∈ Z. |Z(G)|(dim V )  7.2. Induced representations and induced Characters. We follow [9, Chap. 5] and [15, Chap. 7] 7.2.1. Definition of induced representations by tensor product. As I have mentioned before, if G is finite group with G subgroup H and (W, σ) is a representation of H, then the induction of W to G, denoted IndG H W or IndH σ is IndG H W := C[G] ⊗C[H] W, where we again use the correspondence between representations of H and C[H]-modules. This is "extension of scalars" from C[H] to C[G]. Then IndH G W is a C[G]-module and in particular a representation of G. 7.2.2. Remark. Some properties of induced representations are easy to deduce from this description, because they reduce to basic properties of the tensor product, while for others it is more useful to work with a formula for the character of IndG HW . 7.2.3. Induced class functions. Given f : H → C, define f0 : G →  g

7→

f (g) 0

C if if

g∈H g 6∈ H

In other words f 0 = f 1H , where 1H is the characteristic (or indicator) function of H, with value 1 on H and 0 elsewhere. Assume f is a class function on H. Then the induced class function IndG H f : G → C is given by X 1 (7.2.4) (IndG f 0 (xgx−1 ). H f )(g) = |H| x∈G

37

7.2.5. Remark. Notation: Isaacs writes f G for the induced class function, consistent with his notation χH for restriction from G to H. G 7.2.6. Lemma. The induced class function IndG H f is a class function satisfying IndH f (1) = [G : H]f (1).

7.2.7. Proof. Let y ∈ G. Then −1 IndG )= H f (ygy

1 X 0 1 X 0 f (xygy −1 x−1 ) == f (zgz −1 ) = IndG H f (g), |H| |H| x∈G

z∈G

where we have made the change of variables z = xy and used that x 7→ xy is a bijection G → G. 0 −1 Plugging 1 into the formula defining IndG ) = f 0 (1) = f (1) and there are H f , every term in the sum is f (x1x |G| terms.  7.2.8. Theorem (Frobenius Reciprocity). Let f : H → C and F : G → C be class functions. Then G [f, ResG H F ]H = [IndH f, F ]G .

7.2.9. Proof 1 via the defining formula (7.2.4). The definition (7.2.4) of IndG H f is an H-averaged sum over G, while [, ]G is a G-averaged sum over G. Plugging in the former into the latter gives XX 1 [IndG f 0 (xgx−1 )F (g −1 ). H f, F ]G = |G||H| g∈G x∈G

−1

We make the change of variables t = xgx , because int(x) : g 7→ xgx−1 is an automorphism – in particular a bijection – and this way the inner sum will not depend on its variable: 1 XX 0 f (t)F (t−1 ). [IndG H f, F ]G = |G||H| t∈G x∈G

0

So the inner sum is constant, equal to |G|f (t)F (t

−1

. Hence 1 X 0 [IndG f (t)F (t−1 ). H f, F ]G = |H| t∈G

Finally, the sum over G is the sum over H because the summands for g ∈ G \ H are zero by definition of f 0 . So 1 X 0 1 X 0 f (t)F (t−1 ) = f (t)F (t−1 ) = [f, ResG HF ] |H| |H| t∈G

t∈H

 7.2.10. Corollary. Assume ψ is a character of H. Then IndG H ψ is a character of G. 7.2.11. Proof 1 via induced class functions and Frobenius reciprocity. We P know that a class function F on G (which is not identically zero) is a character if and only if its expression F = [F, χ]χ in the orthonormal basis Irr(G) has all coefficients [F, χ] non-negative integers. Given χ ∈ Irr(G), the restriction ResG H χ is a character of H. So G [ResH χ, ψ]H is a non-negative integer. By Frobenius reciprocity, G [IndG H ψ, χ]G = [ψ, ResH χ]H ,

so the coefficients of IndG It is natural to ask which H ψ are non-negative integers. ψ ∈ Irr(H) are restricted from irreducibles of G. This is a difficult question, but if we are content with "contained in the restriction of an irreducible" rather than "is the restriction of an irreducible", then the answer is always: 7.2.12. Corollary. If ψ ∈ Irr(H), then there exists χ ∈ Irr(G) such that ψ is a constituent of the restriction ResG H χ. G 7.2.13. Proof. Let χ be an irreducible constituent of the induced character IndG H ψ. Then [IndH ψ, χ] ≥ 1. By G G G Frobenius reciprocity 7.2.8, [IndH ψ, χ] = [ψ, ResH χ]. So also [ψ, ResH χ] ≥ 1, meaning that ψ is a constituent of ResG  H χ.

7.2.14. Corollary. Assume G is abelian. Then every linear character of H is the restriction of some linear character of G. 7.2.15. Proof. As in Corollary 7.2.12 and its proof, ψ is a constituent of ResG H χ for some χ ∈ Irr(G). But χ is linear since G is abelian. Hence ψ = ResG χ.  H We have the following application to group theory: 7.2.16. Theorem. Assume that for some prime p the p-Sylow subgroups of G are abelian. Then |G0 ∩ Z(G)| is not divisible by p 38

7.2.17. Proof. If not, then there exists a subgroup H ⊂ G0 ∩ Z(G) of order p. Recall that every p-group is contained in a p-Sylow; let P be a p-Sylow containing H. Let λ be a nontrivial linear character of H. By the corollary there exists µ ∈ Irr(P ) whose restriction to H is λ. Induce µ to G and decompose: X IndG [µ, χ]χ Pµ = χ∈Irr(G)

IndG P µ(1)

Since µ(1) = λ(1) = 1, we have = [G : P ]; this is prime to p since P is a p-Sylow. Therefore at least one summand [µ, χ]χ(1) is prime to p. By Frobenius reciprocity 7.2.8, G [IndG P µ, χ] = [µ, ResP χ] ≥ 1.

Hence P P G [λ, ResG H χ] = [ResH µ, ResH ResP χ] ≥ 1,

so λ is a constituent of ResG H χ. Since H ⊂ Z(G), by Schur’s lemma H acts by a scalar – the central character – in a representation ρ affording χ. Z(G) ⊕χ(1) Z(G) That is ResG ωρ , where ωρ is the central character. Hence λ = ResH ωρ , so that ResG H ρ = ResH H χ = χ(1)λ and G χ(1) ResG . H det χ = det ResH χ = λ We are also assuming H ⊂ G0 . Since every linear character of G factors through G/G0 , every linear character χ(1) = 1. Since λ is is trivial on G0 , in particular on H. We apply this to det χ. So ResG H det χ = 1H . Thus λ ∗ µ, χ]χ(1) is prime nontrivial, its order in X (H) is p. Hence p divides χ(1), contradicting that the summand [IndG P to p.  7.3. Monomial characters and M -groups. It is in many ways simpler to induce a linear character than one of higher degree. ∗ 7.3.1. Definition. A character χ of G is called monomial if χ = IndG H λ is induced from a linear character λ ∈ X (H) of some subgroup H of G.

It is allowed to take H = G, i.e., linear characters of G are monomial. 7.3.2. Definition. We say that G is an M -group if every irreducible character of G is monomial. Two basic theorems about M -groups are: In the negative direction, we shall prove: 7.3.3. Theorem (Taketa). Every M -group is solvable. In the positive direction, we shall probably not prove (we could have reached the prove if we had a few more weeks): 7.3.4. Theorem (Artin). Every nilpotent group is an M -group. In particular, every p-group is an M -group. 7.3.5. Remark. A finite group is nilpotent if and only if it is a direct product of its Sylow subgroups. 7.3.6. Remark. A finite group G is supersolvable if it admits a chain 1 = Gm ⊂ Gm−1 ⊂ · · · ⊂ G1 ⊂ G0 = G such that every Gi is normal in all of G and all successive quotients (also called graded pieces) are cyclic. In particular non-abelian groups of order pq are supersolvable but not nilpotent, where p, q are distinct primes. More generally than Artin’s theorem 7.3.4, every supersolvable group is an M -group (see the remarks following the proof of (6.22) in [9]). Note that if G is merely solvable, then there exists a chain of subgroups Gi such that Gi is normal in Gi−1 for all i and the successive quotients are cyclic (but some Gj might not be normal in G) and there also exists a chain with all Gi normal in G (but the successive quotients might be abelian and not cyclic). The three classes – nilpotent, supersolvable and solvable – are summed up concisely by Serre [15, §8.3]. 7.3.7. Remark. The group SL(2, F3 ) is the smallest example of a solvable group (homework) which is not an M group. It is not nilpotent in accordance with Artin’s theorem (the upper and lower triangular matrices with 1’s on the diagonal give two 3-Sylows). 7.3.8. Remark. Isaacs points out that there is no known characterization of M -groups without using characters. We now work towards proving Taketa’s theorem. 39

7.3.9. Lemma. The kernel of an induced character IndG H ψ is given by \ G ker IndH ψ = x(ker ψ)x−1 x∈G

7.3.10. Remark. Note that Isaacs uses the notation g x := xgx−1 ; it is a common notation for conjugation so it’s good to be aware of it. 7.3.11. Proof of 7.3.9. We have g ∈ ker IndG H ψ if and only if G IndG H ψ(g) = IndH ψ(1)

By definition of the induced class function, this is if and only if X (7.3.12) ψ 0 (xgx−1 ) = |G|ψ(1). x∈G

Since every summand ψ 0 (xgx−1 ) has absolute value |ψ 0 (xgx−1) | ≤ |ψ(xgx−1 )| ≤ ψ(1), we have the equality (7.3.12) if and only if ψ 0 (xgx−1 ) = ψ(1) for all x ∈ G, which means xgx−1 ∈ H (since ψ(1) > 0) and xgx−1 ∈ ker ψ. In turn this is if and only if g ∈ x−1 (ker ψ)x for all x ∈ G which is if and only if g ∈ x(ker ψ)x−1 for all x ∈ G (since x 7→ x−1 is a bijection).  7.3.13. Derived series. Recall the commutator subgroup G0 of a group G (1.7.3). The derived series · · · ⊂ G(i+1) ⊂ G(i) ⊂ G(1) = G0 ⊂ G(0) = G of G is a descending filtration of subgroups G(i) indexed by non-negative integers i ≥ 0, defined recursively by G(i+1) = (G(i) )0 That is, we take successive commutator subgroups. Since the commutator subgroup of G is characteristic in G, and since A characteristic in B normal in C =⇒ A normal in C, all the G(i) are normal in G. One of the many equivalent definitions of "solvable" is that G is solvable if G(m) = {1} for some m ≥ 0. 7.3.14. Theorem. Order the set {χ(1) |χ ∈ Irr(G)} of character degrees as 1 = f1 < f2 < · · · < fr . Assume G is an M -group. If χ ∈ Irr(G) has degree χ(1) = fi , then G(i) ⊂ ker χ. Of course there might be multiple irreducible characters whose degree is a given fj . 7.3.15. Proof. By induction on i. If i = 1, then χ(1) = f1 = 1, so χ is linear. Hence G(1) = G0 ⊂ ker χ. Next assume i > 1. If ψ ∈ Irr(G) satisfies ψ(1) < χ(1), then ψ(1) = fj for some j < i. So i − 1 ≥ j, hence G(i−1) ⊂ G(j) and G(j) ⊂ ker ψ by induction. So G(i−1) ⊂ ker ψ provided that ψ(1) < χ(1). Since G is an M -group, there exists a subgroup H and linear λ ∈ X ∗ (H) such that χ = IndG H λ. By Frobenius reciprocity, [IndG H 1H , 1G ] = [1H , 1H ] = 1. So IndG 1 is not irreducible and all of its constituents ψ satisfy H H G ψ(1) < IndG H 1H (1) = [G : H] = IndH λ(1) = χ(1).

By the previous step, G(i−1) ⊂ ker ψ for all constituents ψ of IndG H 1H . Since this is so for all constituents, also G (i−1) G(i−1) ⊂ ker IndG ⊂ H. H 1H . By Lemma 7.3.9 describing the kernel of an induction, ker IndH 1H ⊂ H. So G Hence G( i) = [G(i−1) , G(i−1) ] ⊂ [H, H] = H 0 . As for i = 1, H 0 ⊂ ker λ because λ is linear. Since G(i) is normal in G, it is stable under all conjugation (=inner automorphisms); hence G(i) ⊂ ker λ implies \ G(i) ⊂ x(ker λ)x−1 . x∈G

Finally, since χ =

IndG H λ,

another application of Lemma 7.3.9 gives \ G(i) ⊂ x(ker λ)x−1 = ker χ. x∈G

 40

7.3.16. Proof of Taketa’s theorem 7.3.3. We claim that G(r) = {1}, where r is the number of distinct character degrees. Indeed, by 7.3.14, one has G(r) ⊂ ker χ for all χ ∈ Irr(G). Now it suffices to recall (3.2.5) that \ ker χ = {1}. χ∈Irr(G)

 8. May 7, 2020. Part I: Frobenius groups and other applications of induction 8.1. More on induced Characters. Let G be a finite group, H a subgroup and f : H → C a class function on H. First note that X X (8.1.1) f 0 (xgx−1 ) = f 0 (x−1 gx) x∈G

x∈G

−1

because inversion x 7→ x is a bijection. So it doesn’t matter if we use xgx−1 or x−1 gx in the definition of the induced class function (7.2.4). 8.1.2. Coset representatives. Let T be a set of representatives for the (left) cosets gH of H in G. (In some texts T is called a transversal for H in G). Given h ∈ H, we have f 0 ((xh)−1 g(xh)) = f 0 (h−1 (x−1 gx)h) = f 0 (x−1 gx) since f is invariant under conjugation by H. Thus all |H| elements of the same left coset xH contribute the same to the right-hand sum (8.1.1). Hence X f 0 (t−1 gt) (8.1.3) IndG H f (g) = t∈T

8.1.4. Connection with permutation representations. As I mentioned before, transitive permutation representations are particularly simple examples of induced representations, namely when we induce the trivial representation. This is easy to verify using the formula 7.2.4 for the induced class function: 8.1.5. Lemma. Let G be a finite group with subgroup H. Then IndG H 1H is the linear representation arising from the action of G on G/H. 8.1.6. Proof. Recall that the permutation character χ associated to the action of G on G/H is given by χ(g) = |Fix(g)| = {xH ∈ G/H |gxH = xH}. Since gxH = H if and only if x

−1

gx ∈ H, using a transversal T and (8.1.3) we see that X IndG 10H (t−1 gt) = |Fix(g)|. H 1H (g) = t∈T

 8.1.7. Remark. As mentioned before, Frobenius reciprocity then gives a one-line proof that [χ, 1G ] = 1, a fact we previously saw using Burnside’s lemma. Indeed, [χ, 1G ] = [IndG H 1H , 1G ] = [1H , 1H ] = 1. So we start to see that while permutation representations are nice, their theory stands no chance against linear representations; it’s like combinatorics versus linear algebra – the former stands no chance. 8.2. Frobenius groups. Frobenius groups, which are defined below, are a class of groups which arise naturally in Galois theory and representation theory. In some sense, Frobenius groups are often solvable, but they also played an important role in the classification of finite simple groups. We will see below how they arise naturally in representation theory. As in the homework, the most elementary way in which Frobenius groups arise is via Jordan’s theorem: 8.2.1. Theorem (C. Jordan). Assume G is a finite group and H is a proper subgroup. Then the number of elements of G conjugate to some element of H is at most [ 0 (8.2.2) H ≤ |G| − [G : H] + 1. H 0 ⊂Class(H) Note that the right-hand side |G| − [G : H] + 1 < |G| since H is proper in G; in particular there exists some element of G not conjugate to any element of H. 41



8.2.3. Proof. Revealed after the homework is due. 8.2.4. Lemma. Let G be a finite group and H < G a proper subgroup. Then the following are equivalent: (a) Equality holds in Jordan’s theorem (8.2.2). (b) For all g ∈ G \ H, one has H ∩ gHg −1 = {1}.



8.2.5. Proof. This is similar to the homework. 8.2.6. Definition.

(a) A pair (G, H) as in Jordan’s theorem 8.2.1 is called a Frobenius pair if the equivalent conditions of 8.2.4 hold. (b) A group G is a Frobenius group if it admits a proper subgroup H such that (G, H) is a Frobenius pair. (c) If (G, H) is a Frobenius pair, then H is called a Frobenius complement of G. (d) If (G, H) is a Frobenius pair, then the set [ (8.2.7) N := G \ (H 0 \ {1}) H 0 ∈Class(H)

consisting of the identity and the [G : H] − 1 elements of G singled out by Jordan’s theorem is called the Frobenius kernel of (G, H). 8.2.8. Remark. Since "not conjugate to an element of H" is stable under conjugation, the set N is stable under conjugation. Therefore, if N is a subgroup of G, it is normal. It is a surprising and nontrivial theorem of Frobenius that N is always a subgroup. There is no known proof which does not use characters. 8.2.9. Theorem (Frobenius). Assume (G, H) is a Frobenius pair. Then the Frobenius kernel N (8.2.7) is a subgroup of G. The subgroup N is normal in G and satisfies N H = G, N ∩ H = {1}. 8.2.10. Remark. In general if K is a subgroup of G then a complement of K in G is a subgroup L satisfying LK = G and L ∩ K = {1}. Hence the terminology "Frobenius complement" in reference to H. One says L is a normal complement of K if L is also normal. So we can rephrase the conclusion of 8.2.9 by saying that N is a normal complement of H in G. For those who have seen semi-direct products this says that G = H n N (where those who pay attention to which side gets the vertical line use the convention that the product "opens up" on the factor which receives the action i.e., N is normal so H acts on N by conjugation). 8.2.11. Idea of the proof. As we will show, Frobenius groups have the pleasant property that every irreducible character ϕ of H extends to an irreducible character of G; in other words every irreducible character of H is the restriction of an irreducible character of G. This almost works simply by inducing ϕ from H to G; but it cannot be that simple. Why? (reveal after asking) 8.2.12. Lemma. Assume (G, H) is a Frobenius pair. Let f : H → C be a class function satisfying f (1) = 0. Then the induced class function IndG H f extends f to G: We have G ResG H IndH f = f.

8.2.13. Proof. It suffices to plug in to the formula (7.2.4) defining the induced class function and use the equivalence 8.2.4. Let h ∈ H a non-identity element. If f 0 (xhx−1 ) 6= 0, then xhx−1 ∈ H; equivalently h ∈ x−1 Hx. So f 0 (xhx−1 ) 6= 0 implies h ∈ H ∩ xHx−1 . By 8.2.4, we conclude that f 0 (xhx−1 ) = 0 for all x 6∈ H. On the other hand, we always have f 0 (xhx−1 ) = f (xhx−1 ) = f (h) for all x ∈ H because f is a class function on H. Thus 1 X 0 1 X IndG f (xhx−1 ) = f (h) = f (h). H f (h) = |H| |H| x∈G

x∈H

IndG Hf

It remains to check that extends f at h = 1. But at 1 one just multiplies the original value f (h) by [G : H]; we are done since f (1) = 0 by assumption.  8.2.14. Theorem. Let (G, H) be a Frobenius pair. If ϕ ∈ Irr(H) is nontrivial and irreducible, then G ϕ˜ := IndG H ϕ + ϕ(1)(1G − IndH 1H )

is an irreducible character of G extending ϕ. 42

8.2.15. Proof. Roughly we translate back and forth from irreducible characters and class functions f satisfying f (1) = 0 by adding and subtracting multiples of the trivial representation. In this vein, to get a class function which vanishes at 1 from ϕ, set f := ϕ − ϕ(1)1H . Applying Frobenius reciprocity 7.2.8 followed by the extension lemma 8.2.12 gives G G [IndG H f, IndH f ]G = [f, ResH f ]H = [f, f ]H .

Since ϕ is nontrivial, [ϕ, 1H ] = 0. So [f, f ] = 1 + ϕ(1)2 . Another application of Frobenius reciprocity gives [IndG H f, 1G ] = [f, 1H ] = −ϕ(1). Rearranging the sum defining ϕ˜ as ϕ˜ = IndG H f + ϕ(1)1G and putting together all the pairing computations gives [ϕ, ˜ ϕ] ˜ = 1, 2

because the two self-pairing terms give 1 + 2ϕ(1) and the two cross terms give −2ϕ(1)2 . Since ϕ˜ is defined as a Z-linear combination of characters, norm-squared equal 1 implies cϕ˜ is an irreducible character with c = 1 or c = −1. We confirm that c = +1 by checking that the value of ϕ˜ at 1 is strictly positive: ϕ(1) ˜ = ϕ(1) ≥ 1 since

IndG H f (1)

= 0.



8.2.16. Lemma. Let (G, H) be a Frobenius pair with Frobenius kernel N . If M is a normal subgroup of G and M ∩ H = {1}, then M ⊂ N 8.2.17. Proof. Since M is normal in G, the assumption M ∩ H = {1} implies that also M ∩ xHx−1 = {1}. So M ⊂ N by definition of N as the identity together with the complement of all conjugates of non-identity elements of H.  8.2.18. Proof of Frobenius’ theorem 8.2.9. Let M=

\

ker ϕ. ˜

ϕ∈Irr(H)\{1H }

Then M is a normal subgroup of G; we claim that M = N . First we show M ⊂ N using the criterion 8.2.16: For all nontrivial ϕ ∈ Irr(H), first 1, x ∈ H implies ϕ(x) = ϕ(x) ˜ and ϕ(1) = ϕ(1); ˜ second x ∈ M implies ϕ(x) ˜ = ϕ(1) ˜ since M ⊂ ker ϕ by definition of M . So x ∈ H ∩ M implies ϕ(x) = ϕ(1). So x lies in the intersection of the kernels of all irreducible characters (since ker 1H = H this intersection is the same with our without 1H ), which is trivial 3.2.5. So x = 1 and M ⊂ N . −1 Conversely, if g ∈ N \ {1}, then IndG 6∈ H by definition of N . Hence ϕ(g) ˜ = ϕ(1) = ϕ(1). ˜ H f (g) = 0 since all xgx So g ∈ ker ϕ˜ for all ϕ. We have shown M = N . By definition of N , we have N ∩H = {1}. Then N H = G follows from |N | = [G : H].  8.3. Deeper results about Frobenius groups. 8.3.1. Theorem (Thompson’s thesis, announced in [16]). Assume N is a Frobenius kernel. Then N is nilpotent. 8.3.2. Theorem. Assume H is a Frobenius complement. Then: (a) [Burnside] The Sylow subgroups of H are either cyclic or generalized quaternion. (b) [Zassenhaus, a proof is given in [8, 46.1]] If H is perfect (i.e., H = H 0 ) then H ∼ = SL(2, F5 ). 8.4. Using characters to produce ‘large’ subgroups. 8.4.1. Theorem (Brauer’s proper subgroup lemma). Let A, B be subgroups of G Assume (a) χ is a character of G which doesn’t contain the trivial character: [χ, 1G ] = 0; G G (b) [ResG A χ, 1A ] + [ResB χ, 1B ] > [ResA∩B χ, 1A∩B ]. Then the subgroup hA, Bi generated by A and B is a proper subgroup of G.  8.4.2. It can be interesting to apply Brauer’s theorem even with A ∩ B = {1}, see 9.3.6. As far as I know, the name I have given to the theorem is not standard (as opposed to ‘Burnside’s pa q b theorem’) 43

8.4.3. Proof of 8.4.1. Let (V, ρ) afford χ. For any subgroup H of G, the subspace V H of fixed points has dimension dim V H = [ResG H χ, 1H ] dim V H because V H ∼ as H-representations. Then = 1⊗ H

V A , V B ⊂ V A∩B and hypothesis (b) gives dim V A + dim V B > dim V A∩B . Hence V A ∩ V B 6= (0). If 0 6= v ∈ V A ∩ V B , then v is fixed by both A and B, hence fixed by the subgroup hA, Bi which A, B generate. But hypothesis (a) says that no nonzero vector is fixed by all of G. Hence hA, Bi < G is proper.  9. May 7, 2020. Part II: Every simple group of order 360 is isomorphic to A6 9.1. Review & Motivation. 9.1.1. Review: Every simple group of order 60 is isomorphic to A5 . Recall that in Abstract Algebra we deduced from the Sylow theorems that every simple group of order 60 is isomorphic to A5 . Recall that after several steps one eventually shows that a simple group G of order 60 has a subgroup H order 12, by showing that such H is given by the normalizer of a 2-Sylow in G (the trickiest step is perhaps to rule out the number of 2-Sylows being 15, which is done by considering the normalizer of a nontrivial intersection). Then we conclude by looking at the permutation representation of G on G/H, since |G/H| = 5. 9.1.2. Every simple group of order 168 is isomorphic to GL(3, F2 ). In [2, §6.2, p. 207], a geometric argument is given to show that every simple group G of order 168 is isomorphic to GL(3, F2 ). Using local group theory (Sylow’s theorems and studying normalizers of p-subgroups without characters) one shows that a simple group G of order 168 has two conjugacy P, L classes of Klein 4-subgroups (i.e., isomorphic to Z/2×Z/2); both containing |P| = L | = 7 Klein 4-subgroups. Further U ∈ P normalizes V ∈ L if and only if V normalizes U . The geometric idea is then to define a "finite geometry" consisting of points and lines, where P is the set of points/vertices, L is the set of lines and a point U ∈ P is incident to a line L if U normalizes V . The resulting finite geometry F is the projective plane of order 2 called the Fano plane (depicted on p. 210 of loc. cit.). Still using only local group theory, one finds that every line is incident to 3 points and every point is incident to 3 lines; every two lines determine a unique point and every two points determine a unique line. It follows that |Aut(F )| ≤ 168, since there are at most 7 choices for where to send a point U1 , at most 6 for a second point U2 and at most 4 for a third point U3 not lying on the unique line incident to U1 and U2 ; 7 · 6 · 4 = 168. If G is simple of order 168, then G acts by conjugation on P and L preserving incidence, i.e., G acts on F . Since G is simple, we get G ,→ Aut(F ), so G ∼ = Aut(F ) as the orders are sandwhiched. Finally, one can use Sylow arguments, coupled with the fact that the upper and lower triangular matrices in GL(3, F2 ) are two 2-Sylows, to check that GL(3, F2 ) is simple. 9.1.3. Remark. In particular, once we have the conjugacy class P we know that G simple of order 168 embeds into S7 ; we can also see this more simply by showing that the number of 3-Sylows is 7. However, GL(3, F2 ) is rather small inside S7 ; the lines L provide additional structure which G must preserve besides permuting the 7 points P. 9.1.4. The next case: A6 . The next order which admits a nonabelian simple group is 360; we know that A6 is simple of order 360. The uniqueness statement for orders 60 and 168 generalizes: 9.1.5. Theorem. Every simple group of order 360 = 6!/2 is isomorphic to A6 .  9.1.6. There are nonisomorphic finite simple groups which have the same order. For example PSL(3, F4 ), A8 and PSL(4, F2 ) are all simple of order 20160 and A8 ∼ = PSL(4, F2 ) but A8 6∼ = PSL(3, F4 ) (that the last two are not  isomorphic is elementary and explained in the book by Rotman [12, Th. 8.24]. ( The claimed 3 × 3 companion matrix A in the proof in [12, Th. 8.24] is wrong; what is displayed is rather a 3 × 3 Jordan block. But claimed conclusion is true for both a Jordan block and a companion matrix, so the rest of the proof is unaffected and correct.) 9.1.7. Difficulty. As for A5 , we will be done if we can show that a simple group G of order 360 has a subgroup R of order 60 (= of index 6). However, what does a subgroup of A6 of order 60 look like? It is isomorphic to A5 . By contrast with H ∼ = A4 in A5 , now the subgroup we seek is itself simple. So we cannot produce R as the normalizer of another subgroup. 44

9.1.8. Two approaches. We present two proofs of 9.1.5: One via characters from [9]; the other by generalizing the geometric argument of [2] (which I found when I was a student [5]). In both cases the first step is to work out the local group theory using Sylow’s theorems. 9.2. The local group theory of G simple of order 360. 9.2.1. Theorem. Assume G is simple of order 360. Then (a) Besides the identity, G is composed of • One conjugacy class of 45 involutions, • One class of 90 elements of order 4, • 80 elements of order 3 which break up into two classes, and • One class of 144 elements of order 5. (b) The Sylow numerology of G is |Syl2 (G)| = 45, |Syl3 (G)| = 10 and |Syl5 (G)| = 36. (c) The 3-Sylows are ∼ = Z/3 × Z/3 and intersect pairwise trivially. (d) The 2-Sylows are dihedral; x 7→ CentG (x) and P 7→Z(P) are inverse bijections between the set of involutions in G and Syl2 (G). (e) G has two conjugacy classes of subgroups of order 3, each of size 20 (f ) G has two conjugacy classes of Klein 4-subgroups, each of size 15; every Klein 4-subgroup U of G is the common intersection of 3 Sylow 2-subgroups and NG (U ) ∼ = S4 . 9.2.2. Proof. I work this out in a completely elementary way in [5].  As we follow along in A6 , what are the two classes of order 3 elements in A6 ? For the approach via characters we will also need: 9.2.3. Corollary. Let P ∈ Syl3 (G) and put N := NG (P ). Then N has 6 conjugacy classes. 9.2.4. Proof. Let U ∈ Syl2 (N ). By the local theory theorem 9.2.1, |N | = 36, x ∈ U implies CentN (x) = U and y ∈ P implies CentB (y) = P . Hence N has 6 conjugacy classes (because we know all the centralizers): Besides the identity, 3 are represented by the 3 non-identity elements of U and P \ {1} is a union of two conjugacy classes of elements of order 3.  9.3. Simple groups of order 360 = |A6 | I: Via Characters. 9.3.1. Character theory strategy. We use Brauer’s theorem to successively produce subgroups K ⊂ H ⊂ R such that (a) |K| = 4, |H| = 12, (b) K is characteristic in H (c) G > R > H. It is then easy to rule out |R| = 24 without characters; since a simple group of order 360 cannot have a subgroup of index < 6, we then conclude |R| = 60 as desired. Unfortunately, the process which carries this out has many steps. We highlight the character theory and especially the use of induced characters and leave the Sylow arguments as exercises. Henceforth we let G be simple of order 360. 9.3.2. Remark. As motivation, we can see what happens in A6 and try to reproduce that in G. Ultimately the group R ∼ = A5 is produced as R = hA, Bi where A, B have order 3. It is not hard to see that h(123), (345)i = A5 because the product of the two 3-cycles is the 5-cycle (12345). So A, B are meant to mimic (123) and (345). What makes things complicated is that h(123), (234)i = A4 , so if we choose two distinct subgroups of order 3 poorly we get a subgroup of order 12, not 60; in fact this is how we get H ∼ = A4 in G of order 12. Then, if our first choice is "poor", as they say "If at first you don’t succeed, try, try, try again" and eventually in at most three(!) tries we do get R of order 60. 9.3.3. Irreducible characters of the normalizer N of a 3-Sylow P . By 9.2.3, N has 6 irreducible characters. Since |N/P | = 4, the quotient N/P is abelian. So N 0 ⊂ P and N has at least 4 linear characters. It follows that N has precisely 4 linear characters, N 0 = P and that the remaining two non-linear irreducible characters are both of degree 4 We have 36 = 12 + 12 + 12 + 12 + 42 + 42 9.3.4. Restriction of an irreducible character χ of G to N . Let 1 6= χ ∈ Irr(G). Since G is simple, χ is faithful. G 0 Hence ResG N χ is not a sum of one-dimensional characters, else we would have N = P in its kernel. So ResN χ must contain at least one of the degree 4 irreducible characters; in particular χ(1) ≥ 4. If χ(1) is odd, then ResG N χ has a linear constituent λ; by Frobenius reciprocity this means χ is a constituent of G IndG N λ. So if χ(1) is odd χ(1) ≤ IndN λ(1) = [G : N ] = 10, so χ(1) = 5 or 9 as also χ(1)||G| (4.5.3). 45

9.3.5. The group G has an irreducible character of degree 5. (Question: Following along in A6 , what is such a character? ) Since the sum of the squares of the irreducible character degrees is 360, which is divisible by 4, and since n2 ≡ 1 (mod 4) for all odd n, there are at least three nontrivial irreducible characters of odd degree. Assume G χ ∈ Irr(G) has degree 9. By the previous step, χ appears in IndG N λ for some linear character λ of N . Since IndN λ G has degree [G : N ] = 10, the difference IndN λ − χ is a linear character of G, trivial since G is simple. Hence IndG N λ = χ + 1G . By Frobenius reciprocity, 1 = [IndG N λ, 1G ] = [λ, 1N ]; since λ is linear we conclude λ = 1N . In sum, any irreducible character χ of G of degree 9 is contained in the 10-dimensional IndG N 1N . So there is at most one such χ and G contains at least 2 characters of degree 5. 9.3.6. Applicability of Brauer’s proper subgroup lemma 8.4.1 to pairs of order 3 subgroups. We produce U ⊂ P a subgroup of order 3 such that [ResG U χ, 1U ] ≥ 3. Then Brauer’s proper subgroup lemma shows that any two conjugates of U generate a proper subgroup of G. Indeed, given such U , also [ResG V χ, 1V ] ≥ 3 for any conjugate V of U . So Brauer’s proper subgroup lemma 8.4.1 applies with A = U, B = V and A ∩ B = {1}. 9.3.7. The subgroup H. Put H = hV, U i where V is a conjugate of U which does not centralize U . Then H < G. If |H| = 60, we are done. So assume henceforth |H| < 60. 9.3.8. Ruling out |H| ∈ {24, 30, 36}. Why is |H| = 6 36? Why is |H| = 6 24? Why is |H| = 6 30? 9.3.9. Conclusion: |H| = 12 and H ∼ 6 24, 30, 36. Hence = A4 . We have seen that 12 ≤ |H| < 60 and that |H| = |H| = 12. Since H has two distinct 3-Sylows U, V , one has H ∼ = A4 . 9.3.10. Second application of Brauer’s proper subgroup lemma 8.4.1. Since |ClassG (U )| = 20 but H ∼ = A4 and P each have only 4 subgroups of order 3, there exists W conjugate to U with W 6⊂ H and W 6⊂ P ; then also W does not centralize U as CentG (U ) = P . Second try: Put L := hU, W i. ∼ A4 again (use the ruling out 9.3.8). Then L ∩ H = U If |L| = 60, we are done as before. Otherwise |L| = 12 and L = since W ⊂ L but W 6⊂ H. Arguing as for H, one has [ResG L χ, 1L ] = 2. So G G [ResG L χ, 1L ] + [ResH χ, 1H ] = 2 + 2 > 3 = [ResU χ, 1U ].

Third time’s the charm: Put R := hL, Hi. Brauer’s proper subgroup lemma with A = L, B = H and A ∩ B = U implies that H < R < G. By the ruling out 9.3.8, |R| = 6 24, 36. Since |R| > 12, we conclude that |R| = 60. This completes the character-based proof that G∼  = A6 . 9.4. Simple groups of order 360 = |A6 | II: A geometric approach. 9.5. Connection to Artin L-functions and Artin’s conjecture (not required for the course but very cool). When proving that every nilpotent group is an M -group, Artin was motivated by the following connection with number theory and Galois theory: 9.5.1. Galois extensions of Q. Let F/Q be a finite Galois extension of Q. Each such extension can be described as the splitting field of some f (x) ∈ Q[x], that is as the smallest subfield of C obtained by adding to Q all the roots of f ). The Galois group Gal(F/Q) is by definition the group of ring automorphisms mapping 1 to 1 (=field automorphsms) of F (which necessarily fix Q); Gal(F/Q) has order [F : Q] = dimQ F . 9.5.2. Inverse Galois problem. It is a big open conjecture "inverse Galois problem over Q" that every finite group G occurs as some Gal(F/Q). This is known when G is solvable (proved by Shafarevich but with mistakes, then corrected). For more on the inverse Galois problem, see Serre [14]. 46

9.5.3. Galois representations and Artin L-functions. To a representation ρ : Gal(F/Q) → GL(V ), Artin associated a function L(ρ, s) of a complex variable s called the Artin L-function of ρ as an infinite product over all primes (called an Euler product) Y L(ρ, s) = Lp (ρ, s), p −s

where Lp (ρ, s) is a rational function in x = p and has to do with how the extension F/Q behaves relative to p "what happens when we try to reduce it mod p". Artin showed that L(ρ, s) converges to a holomorphic function for all s with sufficiently large real part i.e., all s in some right half-plane. When ρ is linear and nontrivial, Artin showed that L(ρ, s) admits analytic continuation to an entire function on all of C. Together with Artin’s "reciprocity law" these are two of the culminating points of "Class Field Theory", in turn the deepest development in number theory up to the middle of the twentieth century. Roughly, Class Field Theory describes all abelian extensions of a finite extension K/Q in terms of the failure of its ring of integers OK , and certain subrings of it, to be PID’s. 9.5.4. Generalization of the Riemann zeta function. When F = Q and ρ = 1 is trivial, then L(ρ, s) = ζ(s) is the Riemann zeta function, given by ζ(s) =

X

n−s

n≥1

for Re(s) > 1; it admits analytic continuation to a meremorphic function on C with a simple pole at s = 1 and no other poles. The original Euler product discovered by Euler (hence the name in general) is the identity ∞ X

n−s =

n=1

Y (1 − p−s )−1 p

When dim ρ = 1, ρ factors through the abelianization of Gal(F/Q); by Galois theory the abelianization is the Galois group of F ab /Q, the maximal abelian extension of Q contained in F . Therefore we see that Artin’s theorem on one-dimensional ρ is tied to abelian extensions. In an attempt to generalize Class Field Theory beyond abelian extensions, Artin conjectured: 9.5.5. Conjecture (Artin). Assume ρ : Gal(F/Q) → GL(V ) is irreducible and nontrivial. Then L(ρ, s) is entire. Artin’s conjecture was one of the main motivations for the Langlands program – a program which perhaps unifies more branches of math than every seen before (in particular, representation theory, number theory, algebraic geometry and analysis on Lie groups). Still, most cases of the Artin conjecture remain completely open. For example, there are already two-dimensional cases ρ : Gal(F/Q) → GL(2, C) where the conjecture is not known. The projective image of ρ is the image of the composition Gal(F/Q)

ρ π◦ρ

/ GL(2, C) π

&  PGL(2, C), where PGL(2, C) = GL(2, C)/{λI |λ ∈ C× } is the quotient by the center of GL(2, C). A finite subgroup of PGL(2, C) is either cyclic, dihedral or isomorphic to one of A4 , S4 or A5 . So the projective image of a two-dimensional ρ is one of these groups. If the projective image is not A5 , then the conjecture is known by work of Langlands and Tunnel. But if the projective image is A5 ("the icosahedral case") then the conjecture is still open (and in fact, if you prove this, you can send it straight to the Annals). 9.5.6. Compatibility with induction. Artin showed that L(ρ, s) is compatible with induction (here one must state the conjecture more generally for F/K rather than just F/Q). It follows that if we know the conjecture for L(σ, s) Gal(F/Q) for some subgroup H of Gal(F/Q) and we know that ρ = IndH σ, then the conjecture holds for ρ too. In particular, Artin deduced from his theorem 7.3.4: 9.5.7. Corollary. (a) If the image of ρ is nilpotent, then the Artin conjecture is true: L(ρ, s) is entire. (b) If the image of ρ is an M -group, then the Artin conjecture is true for ρ. 47

9.5.8. Remark. The Artin conjecture is not even known for all ρ with solvable image. It is known for 2-dimensional ρ with solvable image due to advances in the Langlands program by Langlands-Tunnel. 10. Summary

Non-commutative semisimple rings/algebras and their modules (Chapter 1)  The group algebra C[G] (Chapters 1-2)  Character basics: Orthogonality relations kernel & centralizer (Chapter 2) w Integrality & divisibility: Burnside theorems pa q b theorem degree divides order (Chapter 3)

 Tensor products tensor generation symmetric & exterior powers power class functions χ(g r ), νj (χ) (Chapter 4)

w Frobenius groups (Chapter 7)

' Induction: Frobenius reciprocity permutation characters M -groups (Chapter 5)  Irr(G) vs. Irr(N ) for normal N in G (Chapter 6)

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(Wushi Goldring) Department of Mathematics, Stockholm University, Stockholm SE-10691, Sweden [email protected]

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