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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?

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  • $\begingroup$ Do you see any relation between the order of $G$ and the order of the subgroups in question? $\endgroup$ – Alexander Konovalov Apr 28 '14 at 19:34
  • $\begingroup$ @AlexanderKonovalov do you mean that 9 divide 36? If you have a solution, i would appreciate if you share. $\endgroup$ – hehe Apr 28 '14 at 19:37
  • $\begingroup$ not only that 9 divides 36, but more. Perhaps one have to think in terms of looking at prime factors of 36 and 9, and that should lead you to the theorem(s) that may be applied here. $\endgroup$ – Alexander Konovalov Apr 28 '14 at 19:41
  • $\begingroup$ @AlexanderKonovalov, well i know that all subgroups of order 9 are abelian, i also know that they all contain an element of order 3 and i also know they are Sylow p-subgroups... i dont see how i can use this, any tips? $\endgroup$ – hehe Apr 28 '14 at 19:44
  • $\begingroup$ that's true, and now Sylow theorems allow you to reason about the number of such subgroups. $\endgroup$ – Alexander Konovalov Apr 28 '14 at 19:48
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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.

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