This question already has an answer here:
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.
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:
Once you have done these, your are more or less finished.