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If we have a group $G$, and maximal a prime power $p^k$ divides $|G|$ (meaning that $p^{k+1}$ does not divide $|G|$), then we must have a subgroup $H$ of order $p^k$, by Sylow's first theorem.

Let $n_p$ denote the number of Sylow $p$-subgroups of $G$ of order $q=p^m$ for some $m \leq k$. By Sylow's Third theorem, we must haave $n_p | (|G|/q)$ and $n_p = 1\pmod p$.

Is it also necessarily true that $n_p$ is a product of prime powers congruent to $1\pmod p$?

(i.e. every for prime power $p_2|n_p, p_2 = 1\pmod p$)

Based on the various examples of groups I have seen, this seems to be true. If it is, how can it be proved? If not, what is a counterexample?

J. Linne
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Check $S_5$, which has six subgroups of order $5$.

Chris Sanders
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    I got a downvote earlier for not noticing that the question was about prime powers of $n_p$ instead of prime factors. The example which I used, $S_p$, does work as a counterexample for any prime $p\geq 5$, because of Bertrand's postulate. – Chris Sanders Feb 21 '22 at 18:15
  • $S_p$ has exactly $(p-1)!$ subgroups of order $p$. Now Bertrand's postulate guarantees the existence of a prime $q$ between $\frac{p+1}{2}$ and $p-1$. The factor $q$ can therefore only appear once in the factorisation of $(p-1)!$, and since $q$ is less than $p$, it cannot be $1; \text{mod};{p}$ – Chris Sanders Feb 21 '22 at 18:21
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    or $(p-2)!$ subgroups? $(5-1)!=24 \neq 1 \pmod 5$. – J. Linne Feb 21 '22 at 18:43
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    Oh yes, sorry, I kept making small mistakes. But $(p-2)!$ is the correct final answer, as confirmed by https://math.stackexchange.com/questions/1529038/what-is-the-number-of-sylow-p-subgroups-in-s-p – Chris Sanders Feb 21 '22 at 18:45