# Can one find the size of a Sylow normalizer from the character table?

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

-
Nice question ! – Alexander Chervov Sep 8 '12 at 17:25

## 1 Answer

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 $G/O_p(G)$).

-