I recently came across the following problem:
Suppose $G$ is a group with $\vert G \vert =60$ and $\vert Z(G)\vert$ is divisible by $4$. Show that $G$ is Abelian
My Try: By Lagrange's theorem and the hypothesis $4$ divides $ \vert Z(G)\vert$, $$\vert Z(G)\vert\in \{4,12,20,60\}$$
But the index $[G:Z(G)]$ in any finite group is always cannot be prime, so $\vert Z(G)\vert \neq 12,20$
so $\vert Z(G)\vert$ is either $4$ or $60$. If the latter holds, then $G=Z(G) $ and we are done. If the former holds, then $G/Z(G)$ has order $15$ and any group of order $15$ is cyclic and so the quotient is cyclic and so $G$ is Abelian
Is this argument correct? If not, what I'm doing wrong?
If so, is there any other easy method to approach this?