The following is known about finite groups:
(*) If $G/Z(G)$ is cyclic, then $G$ is abelian.
Proposition: Let $G$ be a nilpotent finite group and $N$ a maximal abelian subgroup of $G$. Then $C_G(N) = N$.
Proof: Assume $N < C_G(N) =: C$. Then $C/N$ is a non-trivial normal subgroup of the nilpotent group $G/N$. Then $$ Z(G/N) \cap C/N \ne 1 $$ as the intersection of the center and a normal subgroup in nilpotent subgroups is always non-trivial. Let $U / N$ a cyclic subgroup of $Z(G/N) \cap C/N$ with $N < U \le G$. Then $U$ is a normal subgroup, and abelian by (*), which contradicts the maximality of $N$.
- I do not understand why $C_G(N)$ is normal?
- As a cyclic subgroup of the center is normal, so is $U/N$ normal, and therefore $U$ is normal (as normality is preserved by inverse images of homomorphism), but why does (*) yields that it is abelian, the quotient is not taken by the center? It must imply somehow that $U/Z(U)$ is cyclic, but I do not see how?