Subgroups of p-groups 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.]
 A: 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.
A: George Glauberman and also Jon Alperin and George Glauberman together have written papers
on this topic in recent years. One example is: "A note on abelian subgroups of p-groups."
Groups St. Andrews 2005. Vol. 2, 445–447, London Math. Soc. Lecture Note Ser., 340, Cambridge
Univ. Press, Cambridge, 2007.
A: A. Yu. Olshanskii proved in 1978
that, for any $n$ and any prime $p$, 
there exists a $p$-group of order at least 
$
p^{{1\over 8}(n^2+4n-8)}
$
having no abelian subgroups of order large than $p^n$.
Together with Miller's estimate $p^{{1\over2}n(n+1)}$ (see @m_t's answer), this gives quadratic upper and lower bounds. 

Similar questions for commutative subalgebras in (associative and Lie) algebras was studied by M.V.Milentyeva. The estimates are also quadratic in this case. 
