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.



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> 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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Many p,q-Sylow subgroups
The answer is **no**: see Corollary 1.3 in



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> Robert M. Guralnick; Gunter Malle; Gabriel Navarro, *[Self-normalizing Sylow subgroups](http://www.ams.org/journals/proc/2004-132-04/S0002-9939-03-07161-2/home.html)*, Proc. Amer. Math. Soc. **132** (2004), 973-979. 
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