Theorem 9.3.1 in Hall's group theory says: Let $G$ be a solvable group and $|G|=m\cdot n$, where $% m=p_{1}^{\alpha _{1}}\cdot \cdot \cdot p_{r}^{\alpha _{r}}$, $(m,n)=1$. Let $% \pi =\{p_{1},...,p_{r}\}$ and $h_{m}$ be the number of $\pi -$Hall subgroups of $G$. Then $h_{m}=q_{1}^{\beta _{1}}\cdot \cdot \cdot q_{s}^{\beta _{s}}$ satisfies the following condition for all $i\in \{1,2,...,s\}$.

$% q_{i}^{\beta _{i}}\equiv 1$ (mod $p_{j}$), for some $p_{j}$.

This question arises now that if we replace assumption solvable group with $p$-solvable group whether again Theorem is true. In the other words: Is it true the following claim? Or is there any counterexample for the claim?

Let $G$ be a $p$-solvable group and $|G|=p^{\alpha }\cdot n$ such that $(p^{\alpha },n)=1$($p\neq 2$). Let $% h_{m}$ be the number of Sylow $p-$subgroups of $G$. Then $h_{m}=q_{1}^{\beta _{1}}\cdot \cdot \cdot q_{s}^{\beta _{s}}$ satisfies the following condition for all $i\in \{1,2,...,s\}$.

$q_{i}^{\beta _{i}}\equiv 1$ (mod $p$).

  • $\begingroup$ Doesn't it follow from Zassenhaus' theorem for every finite group $G$? $\endgroup$ – Mark Sapir Aug 4 '18 at 14:26
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    $\begingroup$ @MarkSapir: Not sure what you intend to mean here. The number of Sylow $5$-subgroups of $A_{5}$ is $6 = 2 \times 3$, but neither $2$ nor $3$ is congruent to $1$ (mod $5$). $\endgroup$ – Geoff Robinson Aug 4 '18 at 18:59
  • $\begingroup$ @GeoffRobinson: You are right. My (wrong) recollection of Schur-Zassenhaus theorem was that it basically generalizes the Sylow theorem. It is not true for the "number part" of the theorem. So the question makes sense. $\endgroup$ – Mark Sapir Aug 4 '18 at 19:02
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    $\begingroup$ Geoff's reduction argument, plus Hall's original theorem, seem to show that the original theorem is true if "solvable" is replaced by "$\pi$-solvable", where $\pi$ is as in the original theorem. $\endgroup$ – Richard Lyons Aug 5 '18 at 11:52
  • $\begingroup$ @GeoffRobinson: Thanks for your answer. Whether $N_{N}(P)=C_{N}(P)$?Also, please let me know why $|H:N_{H}(P)=|N:C_{N}(P)|$? $\endgroup$ – R. Kh Aug 9 '18 at 6:47

I think the answer is yes. The number of Sylow $p$-subgroups of $G$ is $[G:N_{G}(P)],$ where $P$ is a Sylow $p$-subgroup of $G$. This number is unchanged if we pass to $G/O_{p}(G),$ so we might as well suppose that $O_{p}(G) = 1.$ Then since $G$ is $p$-solvable, we have $O_{p^{\prime}}(G) = N \neq 1.$

The Schur-Zassenhaus theorem is indeed relevant here, because by the Schur-Zassenhaus Theorem, we have $N_{G^{\ast}}(P^{\ast}) = N_{G}(P)N/N \cong N_{G}(P)/N_{N}(P),$ where $G^{\ast} = G/N$ (though in fact I think the Frattini argument would suffice here).

Hence $[G^{\ast} : N_{G^{\ast}}(P^{\ast})] [N:N_{N}(P)] = [G:N_{G}(P)],$ so by induction, it suffices to prove the result for $NP.$

In other words, it suffices to prove the result for $H = PN,$ which is a group with a normal $p$-complement.

But for each prime divisor $q$ of $|N|,$ there is a $P$-invariant Sylow $q$-subgroup $Q$ of $N$ which contains a Sylow $q$-subgroup of $C_{H}(P).$ Now we have $N_{PQ}(P) = PC_{Q}(P),$ so by Sylow's Theorem, applied in $PQ$, we have $[Q:C_{Q}(P)| \equiv 1$ (mod $p$). But $[H:N_{H}(P)] $ is the product of the various $[Q:C_{Q}(P)]$ ( as $q$ runs through prime divisors of $|N|),$ so the result follows. Note that $[H:N_{H}(P)] = [N:C_{N}(P)]$ since $N$ is a normal subgroup of order prime to $p$ and $P$ is a $p$-group.


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