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We know $|PSL(2,p)|=p(p+1)(p-1)/2$. Let $p$ be Mersenne prime (that is $p+1=2^{n}$) and $r$ be prime divisor of $(p-1)/2$.

My question: What is the number of Sylow $r$-subgroups of $PSL(2,p)$?

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1 Answer 1

up vote 4 down vote accepted

Let $G=\operatorname{PSL}(2,q)$ for q an odd prime power. If r is an odd prime not dividing q but dividing $|G|$, then r divides exactly one of $\{q+1,q-1\}$ and a Sylow r-subgroup of G is cyclic with normalizer dihedral of order $q-\epsilon$ ($\epsilon = \pm1$). In particular, the number of Sylow r-subgroups is $q(q+\epsilon)/2$.

The normalizer is a at least this big because it obviously normalizes the subgroup, and it is no bigger, since it is a maximal subgroup.

This does not require q to be a Mersenne prime, or even prime (just odd), and does not require r divide $q-1$ (though in your case this means $\epsilon=1$).

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I was about to start typing an answer based on a calculation showing that that the normalizer of a non-trivial subgroup of the subgroup of the diagonal matrices (such as these subgroups of order $\mid (p-1)$) consists of the monomial matrices, and thus is dihedral of order $2(p-1)/2$. +1 and thanks for saving me the trouble :-) –  Jyrki Lahtonen Jun 12 '12 at 16:02
    
No problem. I thought this was explicitly done in Gorenstein's textbook, but I don't see it. I've definitely seen it clearly presented in a standard textbook somewhere (and would like to add this to the answer). The only reference I know of right now that has it is the paper that describes the Sylow p-subgroups of PSL(n,q) (where we only need the very easy case for r > n, r not dividing q). –  Jack Schmidt Jun 12 '12 at 16:05
    
@JackSchmidt: thank you so much. –  T R Jun 12 '12 at 16:16

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