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? 
 A: 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 $\varphi: H_3 \to \text{Aut}(H_5 H_{13})$?
A: Here is one more way to solve this problem.
By Sylow's theorem, the $5$-Sylow subgroup $P$ and the $13$-Sylow subgroup $Q$ are both normal in $G$. 
Then $G/Q$ is cyclic, since any group of order $15$ is. In particular it is abelian, so $G' \leq Q$. 
This shows that $P \cap G' = \{ 1 \}$. From this it follows that $P$ is central. If $p \in P$ and $g \in G$, we have $p^{-1}g^{-1}pg \in P \cap G'$, so $pg = gp$. 
In general it is true that if $P \trianglelefteq G$ and $P \cap G' = \{1\}$, then $P$ is central.
A: You have correctly pointed out that $G$ contains a unique (and hence normal) 5-sylow subgroup, $N_5.$ Consider the action of $G$ on $N_5$ given by conjugation. If this is the trivial action, then $N_5$ is central in $G.$
Consider the  homomorphism $\phi: G \rightarrow \mathrm{Aut}(N_5)$ given by sending $g$ to the conjugation by $g$ map. As  $(|\mathrm{Aut}(N_5)|,|G|) = 1,$ the map $\phi$ is trivial. It follows $G$ acts trivially on $N_5.$ 
