# Number of $p$-subgroups of finite group

Assume we have a finite group $$G$$, $$|G|=p^cm$$ where $$p\not\mid m$$ is prime. Fix a subgroup $$H$$ of order $$p^a$$ and a number $$a\le b\le c$$. Prove that the number of subgroups of order $$p^b$$ which contain the subgroup $$H$$ is $$\equiv1\pmod p$$.

I tried to use Wielandt's proof of Sylow's theorem as in Wielandt's proof of Sylow's theorem., but if we set $$S=\{A\subset G, |A|=p^b,H\subset A\}$$, we can't have $$G$$ act on $$S$$ by right multiplication. How can I prove this statement?

• I have not thought this through, but you could try replacing the set of group elements in Wielandt's proof by the set $C$ of $p^{c-a}m$ right cosets of $H$ in $G$, and the set $S$ by the set of subsets of $C$ of size $p^{b-a}$. – Derek Holt Nov 8 '18 at 8:06
• See Theorem 7.9 in math.uconn.edu/~kconrad/blurbs/grouptheory/transitive.pdf. – KCd Nov 22 '18 at 11:28

I suspect that this is something well-known, but I am not aware of any specific references. Anyway, here is a proof.

Theorem. Let $$H$$ be a $$p$$-subgroup of a group $$G$$, and suppose $$p^t \geq |H|$$. Then the number of subgroups $$X \leq G$$ such that $$H \leq X$$ and $$|X|=p^t$$ is congruent to $$1 \bmod p$$.

Lemma 1. The theorem is true in the case where $$p^t$$ is the order of a Sylow $$p$$-subgroup of $$G$$.

Proof. Let $$H \leq P \in \operatorname{Syl}_p(G)$$ and induct on $$|P:H|$$. The result is trivial if $$H=P$$, so we can assume that $$H < P$$ and we let $$N = N_P(H)$$. Conjugation by $$N$$ permutes the Sylow $$p$$-subgroups that contain $$H$$. If $$S$$ is one of these and $$N \nleq S$$, then $$N$$ does not normalise $$S$$, and so the $$N$$-orbit of $$S$$ has size divisible by $$p$$. Also, all members of this orbit fail to contain $$N$$, so we need only count the Sylow $$p$$-subgroups of $$G$$ that contain $$N$$, and since $$N > H$$, this number is congruent to $$1 \bmod p$$ by the inductive hypothesis.

Lemma 2. The theorem is true in the case where $$G$$ is a $$p$$-group.

Proof. Induct on $$|G:H|$$. The result is trivial if $$H=G$$ so we can assume that $$H. We can also assume that $$p^t < |G|$$. Let $$\mathcal{X}$$ and $$\mathcal{M}$$, respectively, be the set of subgroups of order $$p^t$$ that contain $$H$$ and the set of maximal subgroups that contain $$H$$. We count in two ways the number $$N$$ of ordered pairs $$(X,M)$$ with $$X \in \mathcal{X}$$, $$M \in \mathcal{M}$$ and $$X \leq M$$.

First, $$N$$ is the sum over $$X \in \mathcal{X}$$ of the number of maximal subgroups that contain $$X$$. Also, $$N$$ is the sum over $$M \in \mathcal{M}$$ of the number of members of $$\mathcal{X}$$ that are contained in $$M$$. It is easy to see that the number of maximal subgroups of $$G$$ that contain any given proper subgroup of $$G$$ is congruent to $$1 \bmod p$$. The first formula for $$N$$ gives $$N \equiv |\mathcal{X}| \bmod p$$ because the number of $$M \in \mathcal{M}$$ containing any given member of $$\mathcal{X}$$ is congruent to $$1 \bmod p$$. By the inductive hypothesis, the number of members of $$\mathcal{X}$$ contained in any given member of $$\mathcal{M}$$ is congruent to $$1$$, so the second formula for $$N$$ gives $$N \equiv |\mathcal{M}| \equiv 1 \bmod p.$$ Thus $$|\mathcal{X}| \equiv N \equiv 1 \bmod p,$$ as wanted.

Proof of Theorem. Let $$\mathcal{X}$$ be the set of subgroups of order $$p^t$$ that contain $$H$$ and let $$\mathcal{P}$$ be the set of $$P \in \operatorname{Syl}_p(G)$$ that contain $$H$$. We count the number $$S$$ of pairs $$(X,P)$$, where $$X \in \mathcal{X}$$, $$P \in \mathcal{P}$$ and $$X \leq P$$.

First, $$S$$ is the sum over $$X \in \mathcal{X}$$ of the number of $$P \in \mathcal{P}$$ such that $$X \leq P$$. Also, $$S$$ is the sum over $$P \in \mathcal{P}$$ of the number of $$X \in \mathcal{X}$$ such that $$X \leq P$$. Since the number of $$P$$ containing a given $$X$$ is congruent to $$1 \bmod p$$ by Lemma $$\mathbf{1}$$, the first formula for $$S$$ yields $$S \equiv |\mathcal{X}| \bmod p$$. The number of $$X$$ contained in a given $$P$$ is congruent to $$1 \bmod p$$ by Lemma $$\mathbf{2}$$, so the second formula for $$S$$ yields $$S \equiv |\mathcal{P}| \bmod p$$.

Also, $$|\mathcal{P}| \equiv 1 \bmod p$$ by Lemma $$\mathbf{1}$$, so we get $$|\mathcal{X}| \equiv S \equiv 1 \bmod p,$$ which concludes our proof.