Group Theory

3 Sylow Theorems

A \(p\)-Sylow subgroup of a group \(G\) is a maximal \(p\)-subgroup, ie. a maximal subgroup that is killed by powers of \(p\) or a maximal subgroup of order \(p^k\).

Theorem 3.1 (Sylow Theorems). Let \(G\) be a finite group with \(|G| = mp^k\) with \(p\nmid m\). Then if \(n_p\) is the number of \(p\)-Sylows, \(n_p \equiv 1 \pmod p,\) and \(n_p | m\). Moreover, \(G\) acts transitively on the set of \(p\)-Sylows via conjugation, and the size of each \(p\)-Sylow is \(p^k\).

Proof. If \(P\) is a \(p\)-Sylow, \(N_G(P)/P\) cannot have any nontrivial \(p\)-groups or else their preimage in \(N_G(P)\) would be an even larger \(p\)-group, so by Cauchy’s Theorem \(p \nmid [N_G(P):P]\) and for any \(p\)-group \(Q\), \(Q\cap N_G(P) = Q\cap P\). Then let \(A\) be the set of all conjugates of \(P\), which are also \(p\)-Sylows. \(P\) acts on this by conjugation, and by our fact about normalizers, there is a unique fixed point, so by Proposition 1.2 all the other orbits have size dividing \(p\), and \(|A| = [G:N_G(P)]\equiv 1 \pmod p\).Then \(p\nmid [G:N_G(P)]\) so \(|P| = p^k\). Finally for any \(p\)-subgroup \(Q\), its action on \(X\) must have a fixed point, so it is in a conjugate of \(P\). \(n_p|m\) as \(n_p = [G:N_G(P)]|[G:P] = m\). □

Corollary 3.2. Let \(G\) be a group of order \(mp^k\) with \(k(p-1)>m\). Then \(G\) is not simple.

Proof. Apply Lemma 2.8 to the existence of \(p\)-Sylows, using the fact that \(p^k \nmid m!\). □