Question on proof that maximal normal abelian subgroup is self-centralising in nilpotent groups

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?
• For the question on $C_G(N)$ see here. – Dietrich Burde Mar 27 '15 at 19:53
• $C_G(N)$ is normal if $N$ is normal. Furthermore in the proof $N$ is a central subgroup of $U$ and $U/N$ is cyclic, hence by (*) the subgroup $U$ must be abelian. – Mikko Korhonen Mar 27 '15 at 19:58
• Okay, now it is clear, by $N < U,C \le G$ the fact $U/N \le C/N$ implies $U \le C$ which means $N \le Z(U)$, as each element of $n$ commutes with all elements from $U$. Further $N \le Z(U)$ implies by second isomorphism theorem that $$U/Z(G) \cong (U/N)/(N/Z(G))$$ and therefore $U/Z(G)$ is a surjective image of $U/N$, and therefore itself cyclic. – StefanH Mar 27 '15 at 20:25
• Without any assumptions on $G$, maximal abelian subgroup $\iff$ self-centralizing. See here: groupprops.subwiki.org/wiki/… – Eric Auld Sep 7 '16 at 18:13