TITLE: Subgroups of p-groups
QUESTION [6 upvotes]: 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.]

REPLY [2 votes]: 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.