Finding the number of subgroups of a finite group

Question

Let G be a finite non-abelian group of order 39. Then find the number of subgroups of order 3 in G? I don't know how to find I. please help

Answer 1

For any group of order $pq$ with primes $p<q$ and $p\mid q-1$ there is exactly one non-abelian group up to isomorphism, namely the semidirect product of $\mathbb{Z}/p$ by $\mathbb{Z}/q$, see this answer. It also shows in the proof how many subgroups there are of order $p=3$, for $(p,q)=(3,13)$, namely $q=13$.

Answer 2

Hints: make sure you can add details and explanations to the following

1) There is one unique subgroup of order $\;13\;$ which is thus normal

2) If there is one unique subgroup of order $\;3\;$ then it is normal and thus $\;G\;$ is the direct product of abelian groups

3) Thus, there must be exactly $\;13\;$ groups of order $\;3\;$ in $\;G\;$ .