A group of order $195$ has an element of order $5$ in its center Let $G$ be a group of order $195=3\cdot5\cdot13$. Show that the center of $G$ has an element of order $5$. 
There are a few theorems we can use here, but I don't seem to be able to put them together quite right. I want to show that the center of $G$ is divisible by the prime number $5$. If this is the case, then we can apply Cauchy's theorem and we are done. 
By Sylow's theorems we get that there are unique $3$-Sylow, $5$-Sylow, and $13$-Sylow subgroups in $G$. Since they are of prime order, they are abelian. Furthermore, their intersection is trivial (by a theorem I beleive). Does this then guarantee that $G=ABC$ and that $G$ is abelian? 

Hint: There are unique, hence normal, $5$- and $13$-Sylows. Their internal direct product is thus normal and has complementary subgroup equal to one of the $3$-Sylows, so $G$ is a semidirect product of $H_5 H_{13}$ and $H_3$, where $H_p$ denotes a $p$-Sylow (not necessarily unique). What can you say about $\varphi: H_3 \to \text{Aut}(H_5 H_{13})$?
