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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?

Olexandr Konovalov
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  • Do you see any relation between the order of $G$ and the order of the subgroups in question? – Olexandr Konovalov Apr 28 '14 at 19:34
  • @AlexanderKonovalov do you mean that 9 divide 36? If you have a solution, i would appreciate if you share. – hehe Apr 28 '14 at 19:37
  • 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. – Olexandr Konovalov Apr 28 '14 at 19:41
  • @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? – hehe Apr 28 '14 at 19:44
  • that's true, and now Sylow theorems allow you to reason about the number of such subgroups. – Olexandr Konovalov Apr 28 '14 at 19:48
  • @AlexanderKonovalov , i have tried going in that direction, im really stuck... – hehe Apr 28 '14 at 19:54
  • Could you say anything about the possible number of Sylow 9-subgroups? – Olexandr Konovalov Apr 28 '14 at 20:01
  • @AlexanderKonovalov, No i cant :( I can see that every subgroup of order 9 itself contains a subgroup of order 3, i dont get further than this... – hehe Apr 28 '14 at 20:06
  • @hehe What does the third Sylow Theorem tell you about how many Sylow subgroups of order 9 there are? – rogerl Apr 28 '14 at 20:09
  • But if you know Sylow p-subgroups, then you should be able to use Sylow theorems to limit possible numbers of Sylow 3-subgroups (sorry, made a typo calling them "9-subgroups" above - of course these are "3-subgroups") to some fixed case(s) which then perhaps can be dealt with on a per case basis... – Olexandr Konovalov Apr 28 '14 at 20:09
  • @AlexanderKonovalov I'm not so sure that simple counting is enough here... which could be because it's 2:30 AM on my end. Possible # of 3-subgroups: By Sylow, it is 1 mod 3 and divides 4, so to get contradiction suppose we have 4 3-subgroups with trivial intersection. This means there are 1+8+8+8+8 = 33 elements accounted for, but there's no contradiction yet since a priori we could have a 2-subgroup intersecting those 33 elts trivially. So it seems we need more tools than just Sylow's theorems. (Unless I'm missing something obvious) – Chris Brooks Apr 29 '14 at 08:38
  • @DaenerysNaharis: I agree that we need more, Sylow theorems are just the first step to reduce the number of subgroups of order 9 to only two possible cases. I suppose that after that one should re-use some thoughts from this question. – Olexandr Konovalov Apr 30 '14 at 11:43
  • @DaenerysNaharis: just posted an answer - the lead is actually in another related question here. – Olexandr Konovalov May 31 '14 at 20:57

1 Answers1

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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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Olexandr Konovalov
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