most recent 30 from http://mathoverflow.net 2013-06-19T11:06:31Z http://mathoverflow.net/feeds/question/67069 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67069/can-one-find-the-size-of-a-sylow-normalizer-from-the-character-table Jack Schmidt 2011-06-06T18:13:41Z 2011-06-07T06:50:09Z

<blockquote> <p>Is the size of the normalizer of a Sylow <em>p</em>-subgroup determined by the ordinary character table of the group?</p> </blockquote> <p>And if so, how does one calculate it?</p> <p>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 <em>too</em> hard, that is, the character table does not determine the Sylow normalizer's order. If it helps to prove you <em>can</em> find the order, then I am happy to assume one also knows the power maps (and so element orders).</p> <blockquote> <p>Isaacs, I. M.; Navarro, Gabriel. "Character tables and Sylow normalizers." Arch. Math. (Basel) 78 (2002), no. 6, 430–434. MR<a href="http://www.ams.org/mathscinet-getitem?mr=1921731" rel="nofollow">1921731</a> DOI:<a href="http://dx.doi.org/10.1007/s00013-002-8267-4" rel="nofollow">10.1007/s00013-002-8267-4</a></p> </blockquote>

http://mathoverflow.net/questions/67069/can-one-find-the-size-of-a-sylow-normalizer-from-the-character-table/67081#67081 Geoff Robinson 2011-06-06T19:40:28Z 2011-06-07T06:50:09Z

<p>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 $G/O_p(G)$).</p>