Classify groups of order 2015

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

• Is this a question from an on-going contest? – Joel Reyes Noche Dec 9 '15 at 3:18
• $31\equiv 1\pmod5$, so $C_{31}$ has an automorphism of order five (namely $c\mapsto c^2$). It sounds like you know how to use that to construct a non-trivial semi-direct product. There are no such automorphism of $C_{13}$, and the other automorphism of $C_{31}$ of order five are powers of the one I gave, so... – Jyrki Lahtonen Dec 9 '15 at 5:56
• Oh god I hope they don't ask this question in year 2048 – Mario Carneiro Dec 10 '15 at 19:30
• @MarioCarneiro I'm kind of hoping they do ;) – pjs36 Dec 10 '15 at 19:55
• Happy New Year!!! This time, a question of the number of groups of order 2017 would most likely be closed very quickly :) – Alexander Konovalov Jan 1 '17 at 14:24

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 $p|q-1|$.