Let $P$ be a $p$-group and $A$ maximal among abelian normal subgroups of $P$. Show that:
1) $A=C_P(A)$.
2) $|P:A|\mid (|A|-1)!$.
1) If $A$ is an abelian normal subgrup of a certain group $G$ then also $C_P(A)$ is. In fact let $x\in C_P(A)$, let $g\in G$ and consider the quantity $$(gxg^{-1})a(gxg^{-1})^{-1}=gx(g^{-1}ag)x^{-1}g^{-1}$$ By normality of $A$: $$gx(g^{-1}ag)x^{-1}g^{-1}=g(g^{-1}ag)xx^{-1}g^{-1}=a$$ So by maximality $A=C_P(A)$
2) This is more complicated: I must use properties of $p$-groups, for example $Z(P)\cap N\neq 1$ for $1<N\triangleleft G$, there are subgroups of all intermediate cardinalities... But i was not able to combine them together with the information given in 1).