# Group of order $108$ not simple using Sylow theorems on Sylow-$2$-subgroups

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 $$n_2=9$$?

• @Dietrich Burde: My question is more specific than mentioned in the link. – Yadati Kiran Mar 4 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 4 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 4 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 $A_9$. – ancientmathematician Mar 4 at 16:13