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Show that there are no nonabelian simple groups with order smaller than $60$. (The methods we have discussed up to this point are sufficient to deal with every order except for $24$, $36$ and $48$. Here is a hint if $|G| = 36$: Suppose that $G$ has distinct Sylow $3$-subgroups $H$ and $K$. What is $|H \cap K|$? What can you say about $|N(H \cap K)|$? Find a nontrivial proper normal subgroup of $G$.)

This is my last group theory exercise from the book "Abstract Algebra" by gregory T. Lee. I have already managed to cover all the cases except those mentioned in the comment of the problem. In this and in this answer they use Burnside's theorems and in others that I have seen, group actions. But this material is not in the book and I would like to do it by following the hint that is in it, as this is an elementary way of doing it. I just need an extra hint to get started, any help would be appreciated.

Hopmaths
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  • Do you see why $\vert H \cap K \vert = 3$ (and not $1$)? Hint: Otherwise, how many elements of $G$ are not an element of some $3$-Sylow subgroup? – Robert Shore Jun 30 '21 at 00:13
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    Do they really get along without group actions? Many concrete groups are defined in terms of an action and I'd also prefer to have actions available when proving e.g. Sylow – Hagen von Eitzen Jun 30 '21 at 00:26
  • @HagenvonEitzen Yes. If you want you can take a look at the book, the book tries to the maximum to work with elementary methods – Hopmaths Jun 30 '21 at 00:38
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    @RobertShore $|H\cap K|$ is either $1, 3$ or $9$. If it is $9$, then $H=K$, so the the $3$-subgroup is unique and hence normal. If it is $1$, then we have $32$ elements and the $2$-subgroup must be unique. That is, there is a normal subgroup. Thus $|H\cap K|=3$. Right? – Hopmaths Jun 30 '21 at 13:19
  • Now $|N(H\cap K)|$ is either $3, 9$ or $36$. If it is $36$ then $|H\cap K|$ is normal. What about $3$ or $9$? – Hopmaths Jun 30 '21 at 13:41
  • I think that in both cases ($ |N(H\cap K)|=3$ or $9$), we have $|N(H\cap K)|=C|(H\cap K)|$ and therefore $|H\cap K|$ is central. Right? – Hopmaths Jun 30 '21 at 13:52
  • Not that. But do you know that a group of order $9$ (more generally, a group of order $p^2$) has to be abelian? – Robert Shore Jun 30 '21 at 15:49
  • @RobertShore Could you please review my answer and tell me exactly where it is missing? – Hopmaths Jun 30 '21 at 16:05

1 Answers1

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This doesn't use the hint, but it's an answer.

Let $\vert G \vert=36=2^2 \cdot 3^2$. For $G$ to have any chance to be simple, we must have $n_3(G)=4$. Let $H_1, H_2, H_3, H_4$ be the $3$-Sylow subgroups of $G$. Then we can define a homomorphism $\varphi:G \to S_4$ via $\varphi(g)(n)=k$, where $g^{-1}H_n g=H_k$. In other words, we let $G$ permute the Sylow subgroups via conjugation.

Consider $\ker(\varphi)$. We know that $\varphi$ has non-trivial image because the Sylow subgroups are not normal, and therefore can't all be fixed by every element of $G$. Thus, $\ker(\varphi) \neq G$. But $36 \gt 24 = \vert S_4 \vert$, so $\varphi$ cannot be injective and it must have non-trivial kernel, which is therefore a non-trivial normal subgroup of $G$.

Robert Shore
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