A Problem about finite group

Question

Let $G$ be a group with $\vert G\vert=60$ and $\vert Z(G)\vert$ is divisible by 4. I wanted to show that $G$ has a normal subgroup of order 5.

I intend to solve it by using sylow's theorem as follows:

Since $\vert G\vert=60=2^2.3.5,$ then $G$ has Sylow 5-subgroups of order 5, and $n_5\in \{1,6\}$. I need help on how to proceed, please.

Answer 1

Hints:

We have that

$$|Z(G)|\in\{4,\,12,\,20,\,60\}\;\implies\;|G/Z(G)|\in\{15,\,5,\,3,\,1\}\implies G/Z(G)\;$$

is cyclic. But no group whatsoever divided by its center can be cyclic non-trivial, so it must be $G/Z(G)=1\iff G\;$ is abelian.

Answer 2

By Sylow III, the number of Sylow-$5$-subgroups $s_5$ divides $2^2\cdot 3$ and is congruent $1$ modulo $5$. Hence it is $1$ or $6$. But every group of order $60$ having more than one Sylow-$5$-subgroup is simple (see Conrad, Consequences of the Sylow Theorems, Theorem $8.3$). This is a contradiction to $|Z(G)|\ge 4$. It follows that $s_5=1$, so that the unique Sylow-$5$-group of order $5$ is normal.