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Dear friends,

I want to know how we can find the number of Sylow subgroups in groups of Lie type? For example in the group PSL(n,q), where $q=p^s$ how we can find the number of Sylow p-subgroups or the number of Sylow r-subgroups where $r$ is in $pi(G)$?

Best regards

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There are two distinct questions here. As pointed out by gauss, the "equicharacteristic" case (when you look at Sylow $p$-subgroups for the defining characteristic $p$) is fairly straightforward, since it's easy to compute the index of a Borel subgroup in a finite group of Lie type. Standard structure theory (in terms of BN-pairs) found in many sources shows that a Borel subgroup is the Sylow normalizer in this case.

For other primes $r$ dividing the group order, it's much harder to make general statements about the number or structure of Sylow $r$-subgroups. Here it's very useful to study a comprehensive summary of properties of the known finite simple groups: Number 3 (1998) in the series of AMS monographs by Gorenstein, Lyons, Solomon The Classification of the Finite Simple Groups. See in particular sections 3.3 and 4.10 for the two types of primes. This volume has lots of other information about the groups of Lie type, including their orders.

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Any Sylow p subgroup P of a group of lie type contained in a Borel subgroup B and if P is a sylow p subgroup in B then B is its normalizer in G Since all Borel subgroups are conjugate, we conclude that the number of sylow subgroups are exactly the number of Borel subgroups of G.

Now for PSL(2,q ) , it is well known that sylow p- subgroups of it are elementary abelian groups. So P can be considered as $ s$ - dimensional vector space over $F_{p}$ where $q= p^{s}$. So the number of -$p^{k}$ -subgroups of P are exactly the number of k- dimensional vector spaces of P.

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    $\begingroup$ I am confused by your second paragraph - in what sense Sylow $p$-subgroups of $PSL_n(p^s)$ are abelian? In $GL_n(p^s)$, a Sylow $p$-subgroup is conjugate to unitriangular matrices... $\endgroup$
    – Vladimir Dotsenko
    Mar 15, 2012 at 15:28
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    $\begingroup$ That is incorrect, and only correct for $n = 2.$ For example, ${\rm SL}(3,2)$ has dihedral Sylow $2$-subgroups of order $8,$ and these are self-normalizing, so there are $21$ of them. $\endgroup$
    – Geoff Robinson
    Mar 15, 2012 at 15:37
  • $\begingroup$ Yes you are right. for n=2 my answer is correct. but in the other case it is incorrect.Thanks for your comments. What cause such mistake is thinking of $PSL_{2}$ as doubly transitive permutation group.In this case the subgroup of it that fixes one point contain unique sylow p-subgroup which is elementary abelian with order $p^{n-1}$ $\endgroup$
    – gauss
    Mar 15, 2012 at 16:20

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