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?
 A: 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 $H<G$. 
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.
