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Show that any group of order $108$ is not simple.

I can show this using Sylow theorems on Sylow-$3$ subgroups. I was not able to completely justify for Sylow-$2$ subgroups though.

In case of Sylow-$2$ subgroups we have the following :

$n_2|27$ and $n_2\equiv1 \mod2\implies n_2=1,3,9,27$, where $n_2$ is the number of distinct Sylow-$2$ subgroups. For the case $n_2=1$ or $n_2=27$, its quite straightforward and the case $n_2=3$ can be shown using extended Cayley theorem.

For $n_2=9$, using extended Cayley theorem $\exists\; \theta:G\to S_9$ a group homomorphism. But I am not able to show that $\ker\theta\neq\{1\}$. Also if I assume $ H$ and $K$ are two Sylow-$2$ subgroups of order $4$, then $H\cap K\trianglelefteq G$ but I cannot show $|H\cap K|\neq1$ necessarily.

Can anyone suggest how to proceed for this case $n_2=9$?

Yadati Kiran
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  • @Dietrich Burde: My question is more specific than mentioned in the link. – Yadati Kiran Mar 04 '19 at 15:42
  • Yes, I saw it. I am searching for another helpful link using $n_2$ instead of $n_3$. But I think, the argument with $n_3$ is shorter, see here. – Dietrich Burde Mar 04 '19 at 15:43
  • @DietrichBurde: I agree completely. But just to have fun and more understanding I was trying with Sylow-$2$ subgroups $n_2$ but unable to argue for the case $n_2=9$. – Yadati Kiran Mar 04 '19 at 15:45
  • I doubt if in your difficult case that you can show there are two Sylow-2-subgroups which intersect non-trivially, so you'll have to look at the 3-structure at some stage. I have in mind the subgroup $\langle (123), (456), (789), (23)(56), (23)(89) \rangle$ of $A_9$. – ancient mathematician Mar 04 '19 at 16:13

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