group $G$ of order $312$. show that G is not simple

Question

I have a group G of order $312$ and I need to show $G$ is not simple.

What I tried :

I know $312 = 2^3\times39$

so, I know that I have an element of order $2$.

does that mean I have a subgroup of order $2$?

If so then I remember that any subgroup of order $2$ is normal

But, I couldn't figure out the reason for that.

Any help will be appreciated.

Answer 1

Just to remove this post from unanswered queue.

Let $n_{13}$ be the number of Sylow $13$-subgroup of $G$. Then by Sylow's Theorem, $n_{13}\equiv 1 \text{(mod }13)$ and $n_{13}$ divides $2^3\cdot 3 = 24$. This implies $n_{13}=1$, so that there is only one Sylow $13$-subgroup, which is consequently normal. The last assertion follows from the fact conjugation preserves the order of a subgroup. So if there is only one subgroup $H$ of order $13$, then for any $g\in G$, we have $|g H g^{-1}| = |H| = 13$, so $g H g^{-1} = H$, i.e. $H$ is normal.