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In a solvable group $G$, Hall's theorem (see, e.g., Th 9.3.1 in M. Hall's The Theory of Groups) implies that the number of Sylow-$p$ subgroups is a product of numbers each of which is congruent to $1$ mod $p$ and divides a chief factor of $G$, which is stronger than the Sylow result that the number is congruent to $1$ mod $p$ and divides the $p'$ part of $|G|$. It seems to me that the stronger result also holds for non-solvable $G$. If $G$ is simple we are done by the Sylow theorem, since $|G|$ is a chief factor. Indeed if all normal subgroups of $G$ have index a power of $p$, we are done by Sylow, since the $p'$ part of $G$ is then contained in a chief factor. Otherwise, let $H$ be a non-trivial normal subgroup with index not a power of $p$, and $P$ a Sylow-$p$ subgroup of $G$ (and thus $PH < G$). A simple argument shows that the number of Sylow-$p$ subgroups of $G$ is the product of the number of Sylow-$p$ subgroups of $PH$ and the number of Sylow-$p$ subgroups of $G/H$, and we are done by induction since the number of Sylow-$p$ subgroups of $G/H$ is a product of $1$ mod $p$ factors that each divide a chief factor of $G/H$, which is also a chief factor of $G$, and the number of Sylow-$p$ subgroups of $PH$ is a product of $1$ mod $p$ factors each of which divides a chief factor of PH--and there is a chief series of $PH$ that refines a chief series of $G$ on its way from $1$ to $H$, and the $1$ mod $p$ factors must divide one of those chief factors, since they can divide the chief factors between $H$ and $PH$, which are all powers of $p$.

The simple argument referenced is just that any Sylow-$p$ subgroup $Q$ of $G$ is contained in a $QH$ that is conjugate to $PH$ by Sylow. If $P$ and $Q$ collapse to the same factor in $G/H$ if and only if $PH$ = $QH$. If $PH\ne QH$ then their Sylow-$p$ subgroups are distinct from each other (and they have the same number since $QH$ and $PH$ are conjugate in $G$). So $G$ has as many Sylow-$p$ subgroups as the number that $PH$ has times the number of distinct Sylow-$p$s in $G/H$.

Does anyone see any holes in this? Has anyone seen this result in the literature?

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Would this help? A Theorem of Marshall Hall:

Theorem (M. Hall, 1967) Let $N \unlhd G$, $P \in Syl_p(G)$, then $n_p(G)=a_pb_pc_p$, where
$a_p = \#Syl_p(G/N)$
$b_p=\#Syl_p(N)$ and
$c_p=\#Syl_p(N_{PN}(P \cap N)/(P \cap N))$.

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