Show that every group of order 1965 is isomorphic to $\mathbb{Z}/393\mathbb{Z} \rtimes \mathbb{Z}/5\mathbb{Z} $

Question

Where do I start in such problem? $1965=3\times 131\times 5.$

I think was able to show $n_3=1$ so $S_3$ is normal and G is not simple. $n_{131}=1$ What else can I do there? Should I split in the abelian and non-abelian cases or something like it? Should I use that to show that G is abelian somehow? Would you know where I can find similar exercises?

Thank you for your time and patience.

Answer 1

So as you mentioned $n_{3}=n_{131}=1$. Hence we get that there exists a normal Hall $5'$-subgroup, which is isomorphic to $C_{393}$. Now you only need to use Schur–Zassenhaus theorem to see that $G\cong C_{393} \rtimes C_{5}$.

You can find Schur–Zassenhaus theorem here: https://en.wikipedia.org/wiki/Schur%E2%80%93Zassenhaus_theorem.