Let $A$ be maximal among abelian normal subgroups of a $p$-group and show that $A=C_P(A)$ Let $P$ be a $p$-group and let $A$ be maximal among abelian normal subgroups of $P$.  Show that $A=C_P(A)$.
This is the second part of a problem in which I successfully proved the following:
Let $P$ be a finite $p$-group and let $U<V$ be normal subgroups of $P$.  Show that there exists $W \triangleleft P$ with $U<W \le V$ and $|W:U|=p$.
I did this by observing that since $U<V$ are normal in $P$, $(V/U) \triangleleft P/U$ and so $(V/U) \cap Z(P/U)$ is nontrivial.  Now suppose that $|V/U|=p$.  Then it easily follows that $U \triangleleft V \triangleleft P$ and $|V:U|=p$.  Now suppose that $|V/U|>p$.  Then choose a subgroup of $(V/U) \cap Z(P/U)$ of order $p$, which is normal (since it is central) in $P/U$.  This subgroup is of the form $W/U$ for some $W<P$.  Then by the Correspondence Theorem, we have $U \triangleleft W \triangleleft P$ and $|W:U|=p$.
I have been told to apply the first part with $U=A$ and $V=C_P(A)$ and show that $W$ is abelian.  I tried using the same strategy as above, i.e. choosing $W$ from $Z(P/U)$.  However, abelian-ness isn't necessarily preserved under the canonical homomorphism from $P$ to $P/U$.  Even if I could obtain such a $W$ I don't see how $W$ abelian implies that $A=C_P(A)$.
I would appreciate a hint to point me in the right direction with this.  Thanks.
 A: This answer is meant to answer Alex's questions rather than the homework question:
Suppose $U$ is abelian, $V=C_P(U)$ and $U<W \leq V$ with $[W:U]=p$ and $W \unlhd V$. You ask how to show $W$ can be chosen to be abelian.
In fact $W$ is always abelian in this case: Since $[W:U]=p$, there is some $w \in W$ with $W=\langle w, U \rangle$. Since $w \in V=C_P(U)$, one has that $w$ commutes with both itself and $U$; similarly $U$ commutes with both itself and $w$. An element that commutes with all generators lies in the center, so $\langle w, U \rangle \leq Z(\langle w, U\rangle)$, that is, $W$ is abelian.
You also asked how $W$ being abelian implies $C_P(U)=U$. The overall structure of the proof is by contradiction: Let $U$ be maximal amongst abelian normal subgroups. Suppose, by way of contradiction, that $U \neq C_P(U)$. Since $U$ is abelian, $U \leq V=C_P(U)$ and since it is not equal we get $U < V$. Now you find the $W$ as above, so that $W$ is abelian and $W \unlhd P$ and $U < W$. However that directlyh contradicts the hypothesis on $U$: it is not maximal amongst normal abelian subgroups if $W$ is a normal abelian subgroup strictly containing it!
A: It is an easy-to-prove fact that if $\;|P|=p^n\;,\;\;p\;$ a prime, then for any $\;0\le k\le n\;$ there exists a normal subgroup of $\;P\;$ of order $\;p^k\;$ (induction + the center of finite $\;p-$groups is non-trivial).
The above proves your part one without problem.
Let's now try the following (BTW, the claim is true for any nilpotent group $\;P\;$ , not only finite $\;p-$ groups): We have the following
$$A\lhd P\implies C_P(A)\lhd N_P(A)=P$$
Suppose $\;A\lneqq C_P(A)\;$ , so by part one there exists a normal subgroup $\;W\;$ s.t. $\;A<W\le C_P(A)\;,\;\;[W:A]\;$ = p .
But in fact $\;A\;$ is central in $\;W\;$ : $\;A\le Z(W)\;$ , so that $\; W/Z(W)\;$ is cyclic and thus $\;W\;$ is abelian, and here you have your contradiction...
A: Since $A$ is abelian, $A \subseteq C_P(A)$. If $A \neq C_P(A)$, then $C_P(A)/A$ would be a nontrivial normal subgroup of the $p$-group $P/A$, hence would intersect the center of $P/A$ nontrivially. Picking an nonidentity element $\overline{a} \in Z(P/A) \cap (C_P(A)/A)$, and lifting back to $P$, gives an element $a \in C_P(A)$, such that $\langle A, a \rangle$ is an abelian normal subgroup of $P$ properly containing $A$, contradicting maximality.
