Let G be group of order $225=3^2 \cdot 5^2$, where the unique (by Sylow's theorems) 5-Sylow subgroup is non cyclic; let's denote this subgroup by $P$. Show that $3$ divides $\left| Z(G) \right|$, where $Z(G)$ is the center of $G$; conclude that if $5$ divides $\left| Z(G) \right|$ as well then $G$ is abelian.
The only thing I was able to notice is that $P$ is abelian non cyclic so, $P\cong \mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}$ but I'm stuck here.
Thanks in advance
Perhaps this analysis helps: assume that $G$ is not abelian and let $P$ be the normal Sylow $5$-subgroup of $G$. Observe that $G=PQ$, for any $Q \in Syl_3(G)$.
Note that since $P$ is abelian, $P \subseteq C_G(P)$ and thus $|G:C_G(P)|$ divides $9$. By the $N/C$-theorem and the fact that $G=N_G(P)$, $G/C_G(P)$ embeds isomorphically into Aut$(P)=$ Aut$(C_5 \times C_5) \cong GL(2,5)$ (here you need the fact that $P$ is not cyclic!). Hence, $|G:C_G(P))| \mid gcd(9,24\cdot 20)=3$.
So either $|G:C_G(P)|=1$, which would mean that $P \subseteq Z(G)$, implying $G$ is abelian, since also $Q$ is abelian and $G=PQ$, a contradiction.
Or $|G:C_G(P)|=3$, yielding $|C_G(P)|=3 \cdot 5^2$. So, $C_G(P)$ has an element $x$ of order $3$, and again, since $G=PQ$ and $Q$ is abelian for any Sylow $3$-subgroup, it follows that $x \in Z(G)$, whence $3 \mid |Z(G)|$.
Now suppose that $5 \mid |Z(G)|$, then $15 \mid |Z(G)|$, and hence $|G:Z(G)| \leq 15$. If $|G:Z(G)| \in \{1,3,5\}$, then $G$ is abelian (using the well-known fact that if for a group $X$, $X/Z(X)$ is cyclic, then $X$ is abelian). But, there is only one type of group of order $15$ and this is $\cong C_{15}$ (try to prove that yourself), hence if $|G:Z(G)|=15$, then again $G$ must be abelian.
In summary: if $G$ is not abelian, then $3 \mid |Z(G)|$ and $5 \nmid |Z(G)|$.