Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is the size of the normalizer of a Sylow p-subgroup determined by the ordinary character table of the group?

And if so, how does one calculate it?

In a solvable group, apparently one can compute the prime divisors of the Sylow normalizers from the character table (Isaacs–Navarro, 2002), but I don't see any discussion of the entire order. I suppose it must be harder to compute the order, and I somewhat hope it is too hard, that is, the character table does not determine the Sylow normalizer's order. If it helps to prove you can find the order, then I am happy to assume one also knows the power maps (and so element orders).

Isaacs, I. M.; Navarro, Gabriel. "Character tables and Sylow normalizers." Arch. Math. (Basel) 78 (2002), no. 6, 430–434. MR1921731 DOI:10.1007/s00013-002-8267-4

share|improve this question
    
Nice question ! –  Alexander Chervov Sep 8 '12 at 17:25
add comment

1 Answer 1

I don't see a full answer to this at present, but here are some thoughts on the $p$-solvable case. I think it is equivalent in that case to the question: given a Sylow $p$-subgroup $P$ of a finite $p$-solvable group $G$, can we determine the order of $N = O_{p'}(C_{G}(P))$ from the character table of $G$? If we can do always do this, then we can find $|N_G(P)|$ by an inductive argument. It is well known that for such $G$, the subgroup $N$ is contained in $O_{p'}(G)=M$, say, and, in fact, $N = M \cap N_{G}(P).$ Since the character table of $G$ contains that of $G/M$, we can work by induction if we can determine $|N|$. However, on the negative side, while it is possible to determine which are the (necessarily $p$-regular) conjugacy classes of $G$ which meet $N$, it may not be so easy to determine $|N|$ from the character table of $G$. But we can see the equivalence of the questions in this case, because if we can determine $|N_G(P)|$ from the character table of $G$ and $|N_{G/M}(MP/M)|$ from the character table of $G/M$, then we can determine $|N| = |M \cap N_G(P)|$ from the character table of $G$. (Added later: I should have said that if $M = 1$ we can calculate $|N_G(P)|$ by working with $G/O_p(G)$).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.