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Let $p$ be a prime divisor of the order of a finite group $G$, and $\alpha$ the greatest power of $p$ dividing $|G|$. Let $n_k$ be the number of subgroups of $G$ of order $p^k$ (so, $n_\alpha$ is the number of $p$-Sylow subgroups of $G$, usually referred to as "$n_p$"). The following three facts are compatible with the subsequent (conjectured) claim:

  1. $n_\alpha\equiv 1\pmod p$, by Sylow III;
  2. $n_k\ge 1$, for every $k=1,\dots,\alpha$: this is, e.g., the Theorem 2.12.1 in Herstein's Topics in algebra;
  3. $n_1\equiv 1\pmod p$: this is a corollary of this result: in fact, in the notation of the link, $p\mid |X|=n_1(p-1)+1=n_1p-(n_1-1)$ if and only if $p\mid n_1-1$ if and only if $n_1\equiv 1\pmod p$.

Claim. $n_k\equiv 1\pmod p$ for every $k=1,\dots,\alpha$.

According to this and this, the claim holds true for the particular cases $|G|=2^33$ and $|G|=2^43$, respectively.

Q: Is the claim true?

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    Good observation! Yes it is true, see M. Isaacs Finite Group Theory, Problem 1C.8 and here: https://math.stackexchange.com/questions/1586060/the-number-of-p-subgroups-of-a-group (see also the link to a paper of Pierre Deligne there, he proved this (among other results ) while being a student). If you cannot prove 1C.8, I can provide a proof later. – Nicky Hekster Jul 17 '22 at 18:48
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    Please do provide a proof, @NickyHekster. You write so well . . . – Shaun Jul 17 '22 at 18:59
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    Will do that first thing in the morning! – Nicky Hekster Jul 17 '22 at 19:06
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    Thank you, @NickyHekster. To be honest, I'm surprised this is true: it would deserve more popularity, IMHO. I'll try to go through the provided link, waiting for your morning proof :) –  Jul 17 '22 at 19:28
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    Yes I agree, it should deserve more attention since it is such a nice result. You won't find it easily in text books. So, excellent that you had the insights and brought it up! – Nicky Hekster Jul 18 '22 at 07:14
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    This question is a duplicate of this one where you can find Wielandt's proof. I am reluctant to close it as a duplicate because Nicky Hester has gone to the trouble of writing out the proof in Isaac's book. – Derek Holt Jul 18 '22 at 09:28

2 Answers2

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Let us prove the following (this is Problem 1C.8 in M.I. Isaacs Finite Group Theory).

Theorem Let $P \in Syl_p(G)$ and $a$ a non-negative integer. Let $$\Gamma=\{ \text{subgroups of $G$ of order } p^a\}$$ $$\Pi=\{ \text{subgroups of $P$ of order } p^a\}.$$ Then $\#\Gamma \equiv \#\Pi$ mod $p$.

Now, before proving this theorem, let's see how it follows that if $p^a$ divides $|G|$ ($p^a$ not necessarily being the largest $p$-power dividing $|G|$), the number of subgroups of order $p^a$ of $G$ is $\equiv 1$ mod $p$.

If ones takes $a$ in the theorem above such that $p^a$ is the maximum power, clearly $\#\Pi=1$ and $\#\Gamma = n_p(G)$, the number of Sylow $p$-subgroups : one recovers the third Sylow theorem. Further, in general, if $P$ is a $p$-group and $p^a \mid |P|$, then the number of subgroups of order $p^a$ is $\equiv 1$ mod $p$, see for proofs here. These two facts combined with the theorem above gives the required result: if $G$ is a finite group and $p^a \mid |G|$, then the number of $p$-subgroups of order $p^a$ is $\equiv 1$ mod $p$.

Now we need a small lemma.

Lemma Let $P \in Syl_p(G)$ and $Q$ a $p$-subgroup of $G$. Then the following are equivalent.
(a) $Q \unlhd P$
(b) $P \subseteq N_G(Q)$.

Proof (a) $\Rightarrow$ (b) is obvious: since $Q$ is normal in $P$, $P$ normalizes $Q$.
(b) $\Rightarrow$ (a). Since $P \in Syl_p(G)$ and $P \subseteq N_G(Q)$, certainly $P \in Syl_p(N_G(Q))$. But of course, $Q \unlhd N_G(Q)$ and, since Sylow subgroups are conjugate, $Q$ is contained in every Sylow $p$-subgroup of $N_G(Q)$. $\square$

Proof of the Theorem: $P$ acts on $\Gamma$ by conjugation, let $\Omega$ be the set of fixed-points. Then $Q \in \Omega$ if and only if $P \subseteq N_G(Q)$, which by the Lemma is equivalent to $Q \unlhd P$. Hence $\Omega=\{Q \unlhd P: |Q|=p^a\}$. By the Orbit-Stabilizer Theorem and the fact that $P$ is a $p$-group, we conclude $\#\Gamma \equiv \#\Omega$ mod $p$.
$P$ also acts on $\Pi$ by conjugation and the fixed points for this action are exactly the elements of the set $\Omega$. So, $\#\Pi \equiv \#\Omega$ mod $p$, yielding $\#\Gamma \equiv \#\Pi$ mod $p$. $\square$

Final observation We now know that if $G$ is a finite group and $p^a \mid |G|$, then the number of $p$-subgroups of order $p^a$ is $\equiv 1$ mod $p$. From this it equally follows that if $G$ is a finite group and $p^a \mid |G|$, then the number of normal $p$-subgroups of order $p^a$ is $\equiv 1$ mod $p$. Can you prove this?

Nicky Hekster
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This is a theorem of Frobenius proved in [F. G. Frobenius, Verallgemeinerung des Sylow’schen Satzes, Sitzungsber. Preuß. Akad. Wiss. 1895 (1895), 981–993]. The most elegant proof is probably the one by Wielandt, see [Huppert, Satz I.7.2] or https://math.stackexchange.com/a/479869/960602.

Brauer Suzuki
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