Square of order of a Sylow p-subgroup in the nonabelian simple groups Is it true that for all Sylow subgroups $P$ of a nonabelian simple group $G$ that $|P|^2 < |G|$?
If $P$ is abelian, this is an easy consequence of Brodkey's theorem (Suppose that a Sylow $p$-subgroup of a finite group $G$ is abelian. Then there exits Sylow $p$-subgroup $P$ and $T$ such that $P\bigcap T=O_{p}(G)$).
 A: The answer is yes, although all the proofs use the classification of finite simple groups.
Your question was a conjecture of Brauer, which was proven in the 
following paper (theorem 3.6 in there).

Kimmerle, Wolfgang; Lyons, Richard; Sandling, Robert; Teague, David N.
  Composition factors from the group ring and Artin's theorem on orders of simple groups. Proc. London Math. Soc. (3) 60 (1990), no. 1, 89–122. 

It was also proven later by Zenkov and Mazurov in this paper:

Zenkov, V. I.; Mazurov, V. D. On the intersection of Sylow subgroups in finite groups. Algebra and Logic 35 (1996), no. 4, 236–240

The main result of Zenkov and Mazurov states that if $G$ is a finite simple group and $p$ is a prime dividing $|G|$, then some pair $P$,$Q$ of Sylow $p$-subgroups of $G$ have trivial intersection. Then $PQ$ is a subset of $G$ with order $|P|^2 < |G|$.
Finally, it turns out that actually something more general is true. Vdovin has proven that if $N$ is a nilpotent subgroup of any nonabelian simple group $G$, then $|N|^2 < |G|$. See this paper:

Vdovin, E. P. Large nilpotent subgroups of finite simple groups. Algebra and Logic 39 (2000), no. 5, 301–312 link to article

