Show that maximal abelian normal subgroup of $p$-group contains the commutator subgroup 
Let $P$ be a $p$-group and $A$ a maximal abelian normal subgroup of $P$. If $|A : C_A(x)| \le p$ for all $x \in P$, then $P' \le A$.

As far as I know I have no idea how to bring the condition about the index of $C_A(x)$ in $A$ into an argument showing that $P' \le A$, or equivalently that $P/A$ is abelian. The only "idea" what might be helpful: As $A \unlhd P$ the group $P$ acts on $A$ by conjugation, hence it acts on the cosets of $C_A(x)$ in $A$.
Any hints?
 A: This seems quite challenging! For $x,y \in P$, it is enough to prove that $[x,y]$ centralizes $A$, or equivalently that the automorphisms of $A$ induced by conjugation by $xy$ and $yx$ are the same, because then the fact that $A$ is maximal abelian implies $C_G(A)=A$, so $[x,y] \in A$.
This is easy if $C_A(x)=C_A(y)$, or if either $x$ or $y$ centralizes $A$. So assume not and let $N = C_A(x) \cap C_A(y)$. Then $A/N$ is elementary abelian of order $p^2$. Let $C_A(x) = \langle a,N \rangle$ and $C_A(y) = \langle b,N \rangle$. If both $x$ and $y$ centralize $A/N$, then again it is easy, so assume not.
Since $x$ and $y$ generate a $p$-group, and their centralizers project onto different subgroups of $A/N$, they cannot both induce nontrivial actions on $A/N$.
So let's assume that $x$ acts nontrivially and $y$ acts trivially on $A/N$. Then we may assume that $a$ is chosen such that $b^x = ab$. Now $y$ does not centralize $a$, so $a^y=ac$ for some $1 \ne c \in N$. But now we find that $xy$ does not centralize any element of $A \setminus N$, contradicting the assumption that $|A:C_A(xy)| \le p$.
