Confused with a statement A corollary at page 91 of the book Group Theory I by M. Suzuki  is as follows:
Let $A$ be an abelian subgroup of a $p$-group $G$. If $A$ is maximal among abelian normal subgroups of $G$, then $A$ satisfies $C_G(A)=A$. In particular, $A$ is maximal among abelian subgroups of $G$. 
I am confused with the last line of the corollary "In particular....", what exactly does it mean. If $A$ is maximal among abelain subgroups of $G$ whether $A$ will be normal. I am confused because the proof uses the normality of $A$.  
 A: It says "maximal among abelian normal", meaning you are looking at abelian normal subgroups only, and then picking a maximal such: one which is not contained in any other abelian normal subgroup.  It is okay for it to be contained in non-abelian subgroups, or even abelian subgroups provided they are not normal.  The result is trying to prove that it is maximal among all abelian subgroups, not just the normal ones.  If it were contained in some other abelian subgroup $B$, then clearly $B\subseteq C_G(A)$.  But if $A=C_G(A)$, we must have $B=A$ and $A$ is maximal among abelian subgroups.
A: A maximal abelian normal subgroup need not be maximal abelian subgroup. Consider $SL(2,p)$, $p>3$. The only normal subgroups of this group are, the center, $1$ and the whole group. We see that the center is "maximal abelian normal subgroup", but it is not "maximal abelian subgroup" since consider the subgroup generated by a non-central element and the center.
The theorem asserts that in $p$-groups (or nilpotent groups), any maximal abelian normal subgroup is also a maximal abelian subgroup. 
