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edited Jun 7 2011 at 6:50 |
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 $G/O_p(G)$).
(Note added later: the later additional proposed argument of a positive answer in the solvable case was flawed, and has been removed).
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edited Jun 6 2011 at 22:58 |
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 $G/O_p(G)$).
(Note added later: the later additional proposed argument of a positive answer in the solvable case was flawed)flawed, and has been removed).
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edited Jun 6 2011 at 22:43 |
Actually, I think this argument can be made into a positive answer for solvable $G$ (but I am slighly diffident). We can pass to the case $O_p(G) = 1$, since the character table of $G$ tells us the character table of $G/O_p(G)$. Now take a minimal (non-trivial) normal subgroup $M$of minimal order (this can be seen from the character table). Let $H \lhd G$ be Note added later: the normal subgroupwith $H/M = O_p(G/M).$ Let $P$ be a Sylow $p$-subgroup proposed argument of $G$. Now $O_p(H) \leq O_p(G)=1$,so that $C_H(M) = M$ by Hall-Higman (or just that $M = F(H)$ in this case). Now set $Q = P \cap H$,a Sylow $p$-subgroup of $H$. Now $C_M(Q) = Z(H)$ (since $M$ is elementary Abelian ).Hence $C_M(Q) \lhd G$. But $M$ is minimal normal in $G$, and is not contained in $Z(H)$,so that $C_M(Q) = 1$. Hence, positive answer in particular, $C_{M}(P) = 1$. Hence $|N_G(P)| = |N_{G/M}(PM)/M)|$since $M$ is a normal subgroup of order prime to $p$, so that $N_G(P) \cap M = C_M(P) =1.$Thus we can calculate $|N_G(P)|$ within the group $G/M$. We now proceed inductively (if $G/M$has a non-trivial normal $p$-subgroup we pass to a strict homomorphic image. If there is nonon-trivial normal $p$-subgroup, we factor out a non-trivial normal subgroup of minimal order). Hence the character table determines $|N_G(P)|$ for solvable $G$ in the sense that there is aninductive algorithm to calculate this order, given the character table of $G$. case was flawed).
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edited Jun 6 2011 at 21:51 |
withing with $G/O_p(G)$). Actually, I think this argument can be made into a positive answer for solvable $G$ (but I am slighly diffident). We can pass to the case $O_p(G) = 1$, since the character table of $G$ tells us the character table of $G/O_p(G)$. Now take a minimal (non-trivial) normal subgroup $M$of minimal order (this can be seen from the character table). Let $H \lhd G$ be the normal subgroupwith $H/M = O_p(G/M).$ Let $P$ be a Sylow $p$-subgroup of $G$. Now $O_p(H) \leq O_p(G)=1$,so that $C_H(M) = M$ by Hall-Higman (or just that $M = F(H)$ in this case). Now set $Q = P \cap H$,a Sylow $p$-subgroup of $H$. Now $C_M(Q) = Z(H)$ (since $M$ is elementary Abelian ).Hence $C_M(Q) \lhd G$. But $M$ is minimal normal in $G$, and is not contained in $Z(H)$,so that $C_M(Q) = 1$. Hence, in particular, $C_{M}(P) = 1$. Hence $|N_G(P)| = |N_{G/M}(PM)/M)|$since $M$ is a normal subgroup of order prime to $p$, so that $N_G(P) \cap M = C_M(P) =1.$Thus we can calculate $|N_G(P)|$ within the group $G/M$. We now proceed inductively (if $G/M$has a non-trivial normal $p$-subgroup we pass to a strict homomorphic image. If there is nonon-trivial normal $p$-subgroup, we factor out a non-trivial normal subgroup of minimal order). Hence the character table determines $|N_G(P)|$ for solvable $G$ in the sense that there is aninductive algorithm to calculate this order, given the character table of $G$.
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edited Jun 6 2011 at 20:00 |
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
withing $G/O_p(G)$).
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answered Jun 6 2011 at 19:40 |
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 $G$.
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