## 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.