Question on Hall's theorem 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 $p$).
 A: 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.
