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This seems to be related to Sylow's theorems, but I have no idea how?

user2277550
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Assume that $|Z(G)|=4$. $Z(G)$ is normal, so the group $G/Z(G)$ has order 15. But every group of order 15 is cyclic, and that means that $G$ is abelian - contradiction. (Because if $G$ is abelian, then $G=Z(G)$)

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35T41
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  • Another point for http://math.stackexchange.com/questions/999247/if-g-zg-is-cyclic-then-g-is-abelian-what-is-the-point. – lhf Sep 23 '16 at 15:43
  • How is saying that "G is abelian" a contradiction? Are all groups of order 60 abelian? – user247327 Sep 23 '16 at 15:59
  • No: if the quotient of a group by its center is cyclic, the group is abelian. Now $G$ abelian implies $Z(G)=G$, hence $Z(G)$ does not have order $4$. – Bernard Sep 23 '16 at 16:06
  • user247327: I edited – 35T41 Sep 23 '16 at 16:14