Prove that any group of order 15 is cyclic. Prove that any group of order $15$ is cyclic.
I know that if order of group is a prime then the group is cyclic, but how to approach such questions?
 A: The easiest way I know how to approach this uses Sylow theory.  Here's a very rough outline:
Let $G$ be a group such that $|G| = 15$.


*

*Show that the group contains normal subgroups of order $3$ and of order $5$.  Let's call them $H$ and $K$ respectively.

*Prove the following fact: If $H$ and $K$ are normal, $H \cap K = \{e\}$, and $G = HK = \{hk : h \in H, k \in K\}$, then $G \cong H \times K$.  
Hints for #2:


*

*Prove that $H \cap K$ is a subgroup of both $H$ and $K$.

*Show that $HK$ is a subgroup of $G$. 


Once you have done these, you are more-or-less finished.
A: Using Sylow's theorem we show that there are subgroups $H$ and $K$ of order 5 and 3 respectively (and the are cyclic of course). Due to 15=3*5 there is only one subgroup of order 5 and only one subgroup of order 3 (Sylow's theorem says that $N_{p} \equiv 1 $ mod $p$, where $N_{p}$ is the number of subgroups with order $p$, $p \in \{3, 5\}$, and $N_{p}$ divides $|G|$ so $N_{p} = 1$). Moreover, these subgroups are normal. That is why $G=K*H$ and cyclic.
