A well known theorem about abelian subgroups of index $p^2$ in $p$-groups is that 


(\*) *if a $p$-group contains abelian subgroup $A$ of index $p^2$, then it contains an abelian normal subgroup $A\_1$ of index $p^2$* ([see this](http://summit.sfu.ca/system/files/iritems1/4186/b13538421.pdf)).


A non-trivial theorem of Alperin asserts that the above theorem is valid if $p^2$ is replaced by $p^3$. Two natural question arise, which I wouldlike to ask here.


* How long can $A$ and $A\_1$ differ in their structure? (in the proof of (\*), the normal subgroup $A\_1$ obtained is by only considering that $A$ is abelian and without knowing its structure.
* The second obvious and natural question is- *if $G$ contains abelian subgroup of index $p^4$, does $G$ contain abelian normal subgroup of index $p^4$?* (and continue question for higher indices)


The second question may be not fully solved, but What is progress on this question? 



Abelian subgroups of maximum order in $p$-Groups
First I note that your assertion about Alperin's theorem is not true for $p=2$. I think the answer to your second question is negative. There is $p$-groups ($p>3$) having abelian subgroups of index $p^{\frac{p+3}{2}}$, and no normal abelian subgroup of that index. In particular, for $p=5$ the analogue of Alperin's theorem for the index $p^4$ is false. Such a counter example can be found in Berkovich's book "Groups of prime power order I" (see §39)


