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In finite $p$-groups, the number of subgroups of order $p^k$ is congruent to $1 \mod p$.

Is it true that the number of normal subgroups of order $p^k$ is congruent to $1 \mod p$?

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Well a conjugacy class of non-normal subgroups in a finite $p$-group has order divisible by $p$. –  Derek Holt Feb 28 '12 at 7:14
    
I am not sure if your first statement is right without $k$ being the maximal power of $p$ in $|G|$. –  user21436 Feb 28 '12 at 10:05
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In any finite group of order divisible by $p^k$ with $p$ prime, the number of subgroups of order $p^k$ is congruent to 1 mod $p$. –  Derek Holt Feb 28 '12 at 10:44

2 Answers 2

Yes.

The simplest way to see this is to apply the following lemma from group actions:

Lemma. Let $G$ be a group of order $p^n$ ($p$ a prime) acting on a finite set $S$. Let $S_0 = \{x\in S\mid gx = x\text{ for all }x\in G\}$. Then $|S_0|\equiv |S|\pmod{p}$.

Proof. An orbit $\overline{s}$ contains exactly one element if and only if $s\in S_0$; orbits $\overline{s}$ with more than one element have $[G:G_{s}]$ elements, and since $G_s\neq G$, then the number of elements is congruent to $0$ modulo $p$. If $\overline{s}_1,\ldots,\overline{s}_t$ are the distinct orbits with more than one element, then $$|S| = |S_0| + |\overline{s}_1|+\cdots+|\overline{s}_t|\equiv |S_0|\pmod{p}.\quad\Box$$

Now apply the lemma to the set $S$ of subgroups of order $p^k$ and $G$ acting on $S$ by conjugation. $S_0$ is precisely the set of normal subgroups of order $p^k$.

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I am sorry in advance for posting the answer since I could not add comment.

So as in the proof of Arturo Magidin, $|S_0|$ is congruent to $|S|$ modulo $p$. But then how can we get that $|S_0|$ is congruent to $1$ modulo $p$? I tried to take $N\in S_0$ and let $N$ act on $S_0$ by conjugation, but then all orbits are of size $1$?

And in fact, if $G$ is a finite group, then the number of subgroups of order $p^k$ in a Sylow $p$-subgroup is congruent to the number of subgroups of order $p^k$ in $G$.

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