If $G$ is a (non-abelian) $p$-group, $|G|=p^n$, $n>3$, then it is elementary that $G$ contains a (normal) abelian subgroup of order $p^2$. It is also true that $G$ necessarily contains a normal abelian subgroup of order $p^3$ (Group Theory – W. R. Scott).
1) What is the largest possible value of $m$ such that any non-abelian group of order $p^n$ contains a normal abelian subgroup of order $p^m$?
2) What is the largest possible value of $m$ such that any non-abelian group of order $p^n$ contains an abelian subgroup of order $p^m$?
[Please suggest references.]
Best Answer
I've been asked to post the following as an answer, although it does not answer either of your questions (i.e. it does not provide the largest $m$).
Here are some suggested references: G A Miller, On the number of abelian subgroups.. in Messenger Math 36 (1906/7). SC Dancs, Abelian subgroups of finite $p$-groups in Trans AMS 169 (1972). Miller shows a group of order $p^n$ has a normal abelian subgroup of order $p^m$ for some $m$ such that $m (m+1)/2 \geq n$. The inequality is correct. Huppert's Endliche Gruppen book is cited there as an alternative proof of Miller's paper (what I remember of that paper is that it is very hard to read).
Edit: a better reference is Zassenhaus's book `The Theory of Groups', IV.3.4. There you find a simple argument for the lower bound above. It's clear from his proof that the lower bound can be improved if you have control of the number of generators of a maximal normal abelian subgroup, for example in the case that the big group is regular.