Every group of order 36 has a normal Sylow subgroup

Prove that if G is a group of order 36, then G has either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup.

Note that 36 = 2^2 \cdot 3^2, so that Sylow’s Theorem forces n_2(G) \in \{1,3,9\} and n_3(G) \in \{1,4\}.

Suppose n_3(G) \neq 1. Then n_3(G) = 4 \not\equiv 1 mod 9, so that by Lemma 13 and this previous exercise, there exist P_3,Q_3 \in \mathsf{Syl}_3(G) such that P_3 \cap Q_3 is nontrivial. Consider now C_G(P_3 \cap Q_3); since P_3 and Q_3 are abelian, \langle P_3,Q_3 \rangle centralizes P_3 \cap Q_3, and moreover we can see by Sylow’s Theorem that G = \langle P_3, Q_3 \rangle. Thus P_3 \cap Q_3 \leq Z(G), and in fact P_3 \cap Q_3 is the intersection of all Sylow 3-subgroups of G. Thus G contains 2 + 6 + 6 + 6 + 6 = 26 elements of 3-power order. Let P_2 \leq G be a Sylow 2-subgroup; P_2 contains three nonidentity elements, whose products with the nonidentity elements in P_3 \cap Q_3 yield 6 elements of order 6 or 12. There are 4 elements left, which necessarily comprise a unique Sylow 2-subgroup.

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