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Appendix: Algebraic Integers and Characters Appendix: Algebraic Integers and Characters In this appendix, we discuss some aspects of character theory that have a different flavor than the material covered in Chapter 6. Log­ ically speaking, this appendix follows Section 15, although here we require a somewhat higher level of algebraic background than in the main part of the book; for instance, some familiarity with Galois theory would be an asset. Throughout, we let G be a finite group. We say that x E C is an algebraic integer if there is some monic polynomial f E Z[X] such that f(x) = O. LEMMA 1. The set of rational numbers that are also algebraic integers is exactly the set of ordinary integers. PROOF. Suppose that alb E Q is an algebraic integer, where a and b are coprime integers with 1 =1= bEN. Then there are integers co, ... ,Cn-l for some n E N such that (ba)n + Cn-l (a)n-lb + ... + Cl (a)b + Co = O. Upon multiplying through by bn, we see that an == 0 (mod b); this contradicts the fact that a and b are coprime. • 180 Appendix: Algebraic Integers and Characters We will make use of the following standard result, whose proof we sketch in the exercises. PROPOSITION 2. The algebraic integers form a subring of C. • The relevance of algebraic integers to the representation theory of finite groups is established by the following basic fact: PROPOSITION 3. Let X be a character of G. Then X(g) is an algebraic integer for any 9 E G. PROOF. Let 9 E G. Then X(g) is a sum of roots of unity by part (iii) of Proposition 14.4. Any root of unity is an algebraic integer; for instance, if w is an nth root of unity, then f(w) = 0, where f(X) = xn -1 E Z[X]. Since the set of algebraic integers is a ring by Proposition 2, it follows that any sum of roots of unity, and in particular X(g), is an algebraic integer. • The next result is necessary for both of our intended applications. LEMMA 4. If X is an irreducible character of G and 9 E G, then IG: Ca(g)IX(g)/X(l) is an algebraic integer. PROOF. Let S be the simple CG-module having character x. Let 9 E G, let K be the conjugacy class of 9 in G, and let 0: E CG be the class sum EXEK x. Consider the map cp: S ---t S defined by cp( s) = o:s for S E S. We observed in the proof of Theorem 14.3 that 0: lies in the center of CG, and from this it follows that cp E Endca(S); therefore by Schur's lemma, there is some A E C such that o:s = AS for all s E S. By taking traces, we now obtain the equation AX(l) = L X(x) = IKlx(g) = IG : Ca(g)lx(g)· xEK Therefore A = IG : Ca(g)IX(g)/X(l). Let r: CG ---t CG be defined by r(z) = zo: for z E CG. It follows from the proof of Lemma 13.11 that r E Endca(CG). Now since S is a simple CG-module, we can view S as being a submodule of CG, and for 0 f: S E S ~ CG we have r(s) = so: = o:s = AS since 0: is a central element. Therefore A is an eigenvalue of r, and so if we let A be the matrix of r with respect to the C-basis G for CG, then we now have det(AI - A) = o. But we see easily that each entry of A is either 0 or 1, and from this it follows that f(X) = det(XI - A) is Appendix: Algebraic Integers and Characters 181 a monic polynomial in X with coefficients in Z. Since f()..) = 0, we conclude that ).. is an algebraic integer. • We can now prove the first main result of this appendix, a fact that was mentioned in Section 14: PROPOSITION 5. Let X be an irreducible character of G. Then X(l) divides IGI· PROOF. Let gl, ... ,gr be a set of conjugacy class representatives of G. We know for each i that IG : Ca(gi)IX(gi)/X(l) and X(gi) are algebraic integers, the former by Lemma 4 and the latter by part (iv) of Proposition 14.4 and Proposition 3. Using row orthogonality, we see that which by Proposition 2 is a rational algebraic integer; the result now follows from Lemma 1. • The goal of the remainder of this appendix is to prove Burnside's theorem on the solvability of groups of order paqb, which was men­ tioned in Section 11. LEMMA 6. Let X be a character of G, let 9 E G, and define , = X(g)/X(l). If, is a non-zero algebraic integer, then 1,1 = 1. PROOF. We see from part (iii) of Proposition 14.4 that x(1h is a sum of X(l) nth roots of unity, where n is the order of g; therefore 1,1 ::; 1. Suppose that 0 < 1,1 < 1, and assume that, is an algebraic integer. Since, is an average of d complex roots of unity, the same will be true of any algebraic conjugate of ,. (Here we are applying Galois theory. An algebraic conjugate of , is the image of , under any automorphism of K that fixes Q, where K is a suitable extension of Q containing ,.) In particular, every conjugate of, has absolute value at most 1, and thus the product of the conjugates of, has ab­ solute value less than 1. But it follows from [25, Theorem 2.5] that this product is, up to sign, a power of the constant term of the min­ imal polynomial of, over Q, a polynomial that by [25, Lemma 2.12] is monic and has integer coefficients. Therefore, the constant term of 182 Appendix: Algebraic Integers and Characters the minimal polynomial of'Y must be zero, which is a contradiction. Hence 'Y cannot be an algebraic integer. • We now prove a classical theorem due to Burnside. THEOREM 7. If G has a conjugacy class of non-trivial prime power order, then G is not simple. PROOF. Suppose that G is simple and that the conjugacy class of 1 # 9 EGis of order pn, where p is prime and n E N. (Observe that G must be non-abelian in this event.) From column orthogonality, we obtain o 1 r r 0= - = -. LXi(g)Xi(l) = (l/p) + LXi(g)Xi(l)/p, p p i=l i=2 where Xl, ... ,Xr are the irreducible characters of G. Since -l/p is by Lemma 1 not an algebraic integer, it follows from Proposition 2 that Xi(g)xi(l)/p is not an algebraic integer for some 2 :S i :S r. As Xi(g) is an algebraic integer, this implies that p f Xi(l) and that Xi(g) # O. Now IG : Ca(g)1 = pn is coprime to Xi(l), so we have alG: Ca(g)1 + bXi(l) = 1 for some a, bE Z. Thus Xi(g)/Xi(l) = alG : Ca(g)IXi(g)/Xi(l) + bXi(g), which by Proposition 3 and Lemma 4 is an algebraic integer, and therefore IXi(g)1 = Xi(l) by Lemma 6. Consequently, we see that 9 E Zi = {x E G Ilxi(X)1 = Xi(l)}. But Zi = 1 by Corollary 15.11, which gives a contradiction. • Finally, we have Burnside's famed paqb theorem: BURNSIDE'S THEOREM. If IGI = paqb, where p and q are primes, then G is solvable. PROOF. We use induction on a + b. If a + b = 1, then G has prime order and hence is solvable by Proposition 11.1. Hence we assume that a + b ~ 2 and that any group of order pr qB is solvable whenever r+s < a+b. Let Q be a Sylow q-subgroup of G. If Q = 1, then G is a p-group and hence is solvable by Corollary 11.5, so we assume that Q # 1, in which case Z(Q) # 1 by Theorem 8.1. Let 1 # 9 E Z(Q). Then as Q ~ Ca(g), we have IG : Ca(g) I = pn for some n :S a. If n = 0, then 9 E Z(G) and hence G is not simple; and if n > 0, then by Proposition 3.13 and Theorem 7 we see that G is not simple. Therefore, G has a non-trivial proper normal subgroup N. Appendix: Algebraic Integers and Characters 183 By induction, both N and G / N are solvable, and hence G is solvable by part (iii) of Proposition 11.3. • We observed on page 100 that there are non-solvable groups of order paqbrc, where p, q, and r are distinct primes. In fact, up to isomorphism there are eight such groups. Six of these groups are projective special linear groups. (Recall that A5 and A6 are, by Ex­ ercises 6.2 and 6.5, isomorphic with projective special linear groups.) The remaining two groups are members of a family of groups called the projective special unitary groups (see [24, p. 245]). EXERCISES 1. For x E C, let Z[x] = {f(x) I f(X) E Z[X]}j this is an abelian subgroup of C. Show that x is an algebraic integer iff Z[x] is finitely generated. 2. (cont.) Show that the set of algebraic integers is a subring of C.
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