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Prove that the number of normal subgroups of order $p^s$ of a finite $p$-group $G$ is congruent to $1$ mod $p$.

I know this result is a weaker version of Sylow's theorems. But without using Sylow's theorems, how can we prove it?

Thanks for any insight.

user111636
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  • Could you elaborate on how this is a weaker version of Sylow? – Tobias Kildetoft Dec 11 '13 at 13:06
  • You could let $G$ act on the set of all subgroups of order $p^s$ (by conjugation), and count the number of fixed points (note that a fixed point is precisely a normal subgroup) – Prahlad Vaidyanathan Dec 11 '13 at 13:11
  • Are you sure that it concerns normal subgroups? Theorem 4.8 in Rotmans 'An Introduction to the Theory of Groups' states it for subgroups (not necessarily normal). – drhab Dec 11 '13 at 13:16
  • @drhab : proving it for normal subgroups would prove it for the set of all subgroups (by the usual fixed point lemma - see my previous comment) – Prahlad Vaidyanathan Dec 11 '13 at 13:19
  • I know theorem 4.8 in Rotman 'An Introduction to the Theory of Groups', it proves the result for the subgroups of $G$. But how can we prove the case normal subgroups? – user111636 Dec 11 '13 at 13:43
  • Much more is true: please read my entry http://math.stackexchange.com/questions/203850/why-is-the-number-of-subgroups-of-a-finite-group-g-of-order-a-fixed-p-power-cong/203894#203894. It contains the answer to your question. – Nicky Hekster Dec 11 '13 at 13:43
  • @Nicky Hekster: Thank you so much – user111636 Dec 11 '13 at 14:41
  • @User111636 - you are very welcome, I hope you appreciate the beauty of all those theorems! – Nicky Hekster Dec 11 '13 at 14:42
  • @Nicky Hekster: Your Remark of your comment on that link looks like the answer of Pralad Vaidyanathan. – user111636 Dec 11 '13 at 15:02
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    Yes, and his $S_0 \subseteq S$ is the set of normal $p$-subgroups on which $G$ acts trivially by conjugation, So the elements of $S_0$ have an orbit of length $1$, the rest of the elements of $S$ (the non-normals) have an orbit length $\geq p$, so they mod out to $0$. Does that answer your question? – Nicky Hekster Dec 11 '13 at 16:10

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Since you know this fact for all subgroups, the normal case is simple. The following lemma is what you need (the proof is almost identical to the proof, using the class equation, that $Z(G) \neq \{e\}$ when $G$ is $p-$ group)

Let $G$ be a $p-$group acting on a set $S$, and let $$ S_0 = \{x \in S : g\cdot x = x \quad\forall g \in G\} $$ Then, $|S| \equiv |S_0|\pmod{p}$

Now, let $S$ be the set of all subgroups of $G$ of order $p^k$, and let $G$ act on $S$ by conjugation. Check that $S_0$ is precisely the set of all normal subgroups of order $p^k$.

Prahlad Vaidyanathan
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  • I see your idea. But can you explain more precise why $S_0$ is the set of all normal subgroups of order $p^k$? – user111636 Dec 11 '13 at 14:09