TITLE: intersection of all subgroups of order 9 in a group of order 36 is non trivial
QUESTION [7 upvotes]: Let $G$ be a group of order 36, prove that all its subgroups of order 9 intersect in a non-trivial subgroup.
I have proven that they intersect in a subgroup, but i cant prove that it contains an element $a \not= e$, can anyone give me a proof for the fact that it is non-trivial?

REPLY [3 votes]: Let $G$ be a group of order $36$. Then by Sylow theorems, the number of subgroups of order $9$ is either $1$ or $4$. Now, a group of order $36$ must have a normal subgroup $N$ of order $3$ or $9$, see e.g. this question. If $N$ has order $9$, we are in the 1st case. If $N$ has order $3$, we have 4 Sylow $3$-subgroups. Since they are all conjugate, and $N$ is normal, $N$ is their intersection.