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What is the number of Sylow 2-subgroups of the group $\mathrm{PSL}(2,q)$?

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Have a look at this: mathoverflow.net/questions/91284/…. They consider the (much more general) problem involving finite groupe of Lie type. –  M Turgeon Jul 13 '12 at 13:41
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When $q$ is a power of $2,$ we have ${\rm PSL}(2,q) = {\rm SL}(2,q)$ and a Sylow $2$-normalizer is a Borel subgroup of order $q(q-1).$ Hence there are $q+1$ Sylow $2$-subgroups as ${\rm SL}(2,q)$ has order $(q-1)q(q+1)$. When $q$ is odd, the order of ${\rm PSL}(2,q)$ is $\frac{q(q-1)(q+1)}{2}.$ A Sylow $2$-subgroup of ${\rm SL}(2,q)$ is (quaternion or) generalized quaternion and a Sylow $2$-subgroup of ${\rm PSL}(2,q)$ is either a Klein $4$-group or a dihedral $2$-group with $8$ or more elements. In all these cases, a Sylow $2$-subgroup of ${\rm SL}(2,q)$ contains its centralizer, and some elementary group theory allows us to conclude that the same is true in ${\rm PSL}(2,q).$ The outer automorphism group of a dihedral $2$-group with $8$ or more elements is a $2$-group. Hence a Sylow $2$-subgroup of ${\rm PSL}(2,q)$ is self-normalizing when $q \equiv \pm 1$ (mod 8), and in that case the number of Sylow $2$-subgroups of ${\rm PSL}(2,q)$ is $q(q^{2}-1)_{2^{\prime}}$ where $n_{2^{\prime}}$ denotes the largest positive odd divisor of the positive integer $n.$ When $q \equiv \pm 3$ (mod 8), then a Sylow $2$-normalizer of ${\rm PSL}(2,q)$ must have order $12$ ( a Sylow $2$-subgroup is a self-centralizing Klein $4$-group, but there must be an element of order $3$ in its normalizer by Burnside's transfer theorem). In this case, the number of Sylow $2$-subgroups of ${\rm PSL}(2,q)$ is $q(\frac{q^{2}-1}{24})$

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@Geoff Robinson: Very nice! thank you very much. –  Sara Jul 13 '12 at 14:30
    
Can you give me an idea on how to prove that the Sylow $2$-subgroup of $\operatorname{SL}(2,q)$ contains its centralizer when $q$ is odd? –  Mikko Korhonen Mar 26 '13 at 21:38
    
@m.k.: When $q$ is odd, a Sylow $2$-subgroup $S$ of ${\rm SL}(2,q)$ is (generalized) quaternion. In particular, it is an (absolutely) irreducible subgroup in the given representation. So it only commutes with scalar matrices by Schur's Lemma. But the only scalar matrices in $ G = {\rm SL}(2,q)$ are $\pm I,$ which are both contained in $S.$ Hence $C_{G}(S) \subseteq S.$ –  Geoff Robinson Mar 27 '13 at 0:08
    
@GeoffRobinson: Thanks! –  Mikko Korhonen Mar 27 '13 at 8:18
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