Abstract Algebra Instructor: Mohamed Omar Lecture - Class Equation/ Intro to Sylow Theorems Oct 27 Math 171 We now define an action of G on itself by g · x = g −1 xg. We proved last class that this is a group action. Definition. The orbits of G under the above action are called the conjugacy classes of G. Example. of σ is 1. Consider σ ∈ Sn . One can prove that if the disjoint cycle decomposition σ = (a1 a2 · · · ak )(b1 b2 · · · bj ) · · · and τ ∈ Sn then τ στ −1 = (τ (a1 ) τ (a2 ) · · · τ (ak ))(τ (b1 ) τ (b2 ) · · · τ (bj )) · · · To see this, observe that if σ(i) = j then τ στ −1 (τ (i)) = τ (σ(τ −1 (τ (i)))) = τ (j). Thus any element conjugate to σ has the same number of cycles of any given length in its disjoint cycle decomposition. It turns out that two elements in Sn are conjugate if and only if they have the same number of cycles of any given length in their disjoint cycle decompositions. For instance, (1 2 4)(3 5) and (1 3 2)(4 5) are conjugate in S5 . 2. When are two matrices in GLn (R) conjugate? A, B ∈ GLn (R) are conjugate if there exists P ∈ GLn (R) such that A = P BP −1 . In other words, A and B are conjugate if and only if they represent the same linear transformation T : Rn → Rn , but under potentially different bases. Observe then that for any x ∈ G, CG (x) = Gx , the stabilizer of x in G under this action. By the Orbit-Stabilizer Theorem, |Gx | = |G|/|Gx| so |Gx| = |G|/|CG (x)|. Now X |G| = |Gx|. x a representative from an orbit of G Observe that any element of Z(G) has orbit size 1 under the above action, so the previous equation implies X |G| = |Z(G)| + |G|/|CG (x)|. x a representative from a conjugacy class of G This is called the Class Equation. Theorem 1 Let p ≥ 2 be a prime. If G is a group and |G| = p2 , then G is abelian. 1 Proof. • If there exists g ∈ G with |g| = p2 , then G ∼ = Z/p2 Z, so G is abelian. • If |Z(G)| = p2 , then G = Z(G) (because Z(G) ≤ G and |G| = |Z(G)| in this case) so G is abelian. Thus we are left to consider when |Z(G)| ∈ {1, p}. • For any x ∈ / Z(G), CG (x) can not be all of G, so [G : CG (x)] is either p or p2 . The class equation tells us it can not be p2 because |Z(G)| ≥ 1, so |G|/|CG (x)| = p. • We now have |G| |{z} divisible by = |Z(G)| + r X i=1 p |G|/|CG (gr )| , | {z } divisible by p so |Z(G)| is divisible by p. From the second bullet point, this implies |Z(G)| = p. • Recall Z(G) G, so we can form the quotient G/Z(G). From the previous part, |G/Z(G)| = p so it is cyclic (because it has prime order). Let Z(G), Z(G)x, Z(G)x2 , . . . , Z(G)xp−1 be the cosets of G/Z(G). Then if a, b ∈ G, there exist y, z ∈ Z(G) and i, j ∈ Z such that a = yxi and b = zxj . But then ab = yxi zxj = yzxi xj = yzxi+j = yzxj xi = zxj yxi = ba, all equalities following from the fact that y, z ∈ Z(G). But this means that for any a, b ∈ G we have ab = ba. Thus, G is abelian. On Assignment 7 we will see that all finite abelian groups are products of cyclic groups, and hence the above theorem tells us that there are two isomorphism classes of groups of order p2 if p is prime: Z/p2 Z and Z/pZ × Z/pZ. This can actually be proven directly by modifying the above proof. Another example of the power of the Class Equation is in its use to prove the following partial converse to Lagrange’s Theorem. Theorem 2 (Cauchy’s Theorem) Let G be a finite abelian group and assume p ≥ 2 is a prime with p | |G|. Then G has an element of order p. 2

© Copyright 2018 ExploreDoc