This question already has an answer here:
- $Z(G)$ is not maximal subgroup of G 1 answer
How to prove that Z(G) is not a maximal subgroup of G, where G is an arbitrary group?
Thanks in advance.
The hint given in the comments is probably the easiest way to do this. Here's an idea for a different solution.
Hint 1: If a maximal subgroup $M$ is normal in $G$, then $G/M$ is cyclic.
Hint 2: If $G/Z(G)$ is cyclic, then $G$ is abelian.