Classify groups of order 2015

Question

How many groups are there of order 2015? Why? I've found that if the group is abelian it is the cyclic group of order 2015. If not, I've tried applying Sylow's theorems and found that there are normal cyclic subgroups of order 13 and 31. I'm not sure how to work out the number of non-isomorphic semi direct products

Answer 1

This is the first idea that came to mind. It probably has an error. I would appreciate any suggestions about how to make this more correct or precise.

Note $2015=5\times 13\times 31$. As you noted, we have normal cyclic subgroups of orders $13$ and $31$. Let $A$ be the group of order $13$ and $B$ be the group of order $31$. We have a map $\phi:A\rightarrow \operatorname{Aut}(B)$. Since $|\operatorname{Aut}(B)|=30$, this must be the trivial action, so $A$ and $B$ commute and they generate a cyclic group of order $13\times 31$. The same kind of argument shows that the group of order $5$ commutes with the group of order $13$. So it remains to see how the group of order $5$ interacts with the group of order $31$.

First, it can commute with the group of order $31$, and we get the cyclic group of order $2015$.

Otherwise, conjugating $B$ by the order $5$ group $C$ induces a nontrivial automorphism. I think you want to show that no matter what automorphism you choose, you always get the same group. This would be similar to the analysis of groups of order $pq$ where $p|q-1|$.