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It is a fact that the symmetric groups have as many 2-Sylow subgroups as possible. More precisely, for all $n \geq 1$, the number of 2-Sylow subgroups in $S_n$ is exactly $n!/2^{\nu_2(n!)}$, which is the index of a 2-Sylow subgroup of $S_n$. This follows from (or, depending on which direction you're coming from, proves) the fact that one (equivalently, all) 2-Sylow subgroup is self normalizing.

It isn't too hard to show that given a prime $p$, there is a family of finite groups $(G_n)$ such that $\nu_p(|G_n|) \rightarrow \infty$ and all the $p$-Sylow subgroups of $G_n$ are self-normalizing.

I want to generalize this to two primes in the obvious way, but I am encountering difficulty. The following would be a good start.

Given distinct primes $p$, $q$, does there exist a finite group $G$ such that $pq \ \Big| \ |G|$ and all $p$-Sylow, all $q$-Sylow subgroups are self-normalizing?

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The answer to your highlighted question is no: see Corollary 1.3 in ams.org/journals/proc/2004-132-04/S0002-9939-03-07161-2/…. (Found through a google search for "self-normalizing sylow p-subgroup") –  Faisal Nov 6 '11 at 4:08
    
Faisal, this is indeed the exact answer to my question! Why don't you post it as an answer so I can confirm it? Thanks! –  Dan Glasscock Nov 6 '11 at 14:32
    
Will do. I left it as a comment in case you wanted to reformulate your question. –  Faisal Nov 6 '11 at 19:39

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up vote 7 down vote accepted

The answer is no: see Corollary 1.3 in

Robert M. Guralnick; Gunter Malle; Gabriel Navarro, Self-normalizing Sylow subgroups, Proc. Amer. Math. Soc. 132 (2004), 973-979.

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