Prove that if is a group of order 36, then has either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup.
Note that , so that Sylow’s Theorem forces and .
Suppose . Then mod 9, so that by Lemma 13 and this previous exercise, there exist such that is nontrivial. Consider now ; since and are abelian, centralizes , and moreover we can see by Sylow’s Theorem that . Thus , and in fact is the intersection of all Sylow 3-subgroups of . Thus contains elements of 3-power order. Let be a Sylow 2-subgroup; contains three nonidentity elements, whose products with the nonidentity elements in yield 6 elements of order 6 or 12. There are 4 elements left, which necessarily comprise a unique Sylow 2-subgroup.