Wielandt's proof of Sylow's theorem. Let $|G|=p^{n}m$, and let $0\leq k\leq n$. Then the number of subgroups of $G$ of order $p^k$ is congruent to $1$ modulo $p$.
I know Wielandt's proof in the case $k=n$, which is standard, but I dont know how to use that proof to prove this above modification. Clearly the set $S$ on which we should make $G$ act by left multiplication has to be the set of all subsets of size $p^k$ and then we know that $\displaystyle \binom{p^{k}r} {p^{k}}$ is congruent to $r \bmod p$. But somehow using the Orbit-Stabilizer theorem I am unable to prove the above modification.
Could someone help me prove the above statement using a modification of Wielandt's proof?
 A: None of the versions of Wielandt's proofs of Sylow's Theorem that I have seen assume that $k=n$. I will cut and paste the proof from some lecture notes I have.
Theorem. Let $G$ be a finite group and let $p^\beta$ divide
$|G|$, where $p$ is prime. Let $k$ be the number of subgroups of $G$ of order
$p^\beta $.  Then $k \equiv 1 \pmod{p} $.
Proof.
Let $|G| = p^\alpha t$ with $p\nmid t$, so $\beta \le \alpha$.
Let $\Omega $ be the set of all subsets of $G$ of order $p^\beta $.
So
$|\Omega | =
\left( \begin{array}{c} p^\alpha t \\ p^\beta \end{array} \right)$.
Let $G$ act on $\Omega $ by right multiplication; i.e. if
$S \in \Omega$, then $S^g = Sg = \{sg \mid s \in S\}$.
Let $\Gamma$ be an orbit of $G^\Omega$.
If $T \in \Gamma$ and $x\in T$ then $1\in Tx^{-1} \in \Gamma$ so there is a set
$S\in \Gamma$ with $1\in S$.
Consider the stabilizer, ${\rm Stab} _G(S)$.
If $g\in {\rm Stab} _G(S)$ then $Sg=S$ so $1g = g \in S$.
Thus ${\rm Stab} _G(S) \subseteq S$.

*

*Suppose that ${\rm Stab} _G(S) = S$, so $S$ is a subgroup of $G$.
Then, by the Orbit-Stabilizer Theorem,
$|\Gamma| = |G|/|{\rm Stab} _G(S)| = p^{\alpha }t/p^{\beta } =
p^{\alpha -\beta }t $ and
$\Gamma$ is the set of right cosets of $S$ in $G$. Thus only one element
of $\Gamma $ is a subgroup.
Conversely, if $T$ is a subgroup of $G$ of order $p^{\beta}$ then
$T^G$ (the orbit of $G^\Omega$ containing $T$) is the set of
right cosets of $T$ in $G$ so has length $p^{\alpha -\beta }t $


*Suppose that ${\rm Stab} _G(S) \ne S$.
Then $|S| > |{\rm Stab} _G(S)|$, so  $|\Gamma | > p^{\alpha -\beta }t $.
Since $|\Gamma |$ divides $|G| = p^{\alpha }t $ we have
$p^{\alpha -\beta +1}$ divides $|\Gamma |$.
So by 1) no element of $\Gamma$ is a subgroup of $G$ in this case.
Hence, there are exactly $k$ orbits whose stabilizer has size $p^\beta $
and these orbits have have length $p^{\alpha -\beta }t $, whereas
those orbits whose stabilizer has size less than $p^\beta $ have length
divisible by  $p^{\alpha -\beta +1} $.
So
$|\Omega | = kp^{\alpha -\beta }t + l p^{\alpha -\beta +1}  $ for some $l$,
and hence $$|\Omega |/p^{\alpha -\beta } = kt + lp \equiv kt \pmod{p}.$$
Since $p \nmid t$, there is a unique $u \in \{1,\ldots,p-1\}$ with
$ut \equiv 1 \pmod{p}$, and multiplying by $u$ gives
$$k \equiv |\Omega |u/p^{\alpha -\beta } \!\!\!\pmod{p} \equiv
 \left( \begin{array}{c} p^\alpha t \\ p^\beta \end{array} \right)u
/p^{\alpha -\beta }\!\!\! \pmod{p}.$$
It is possible to prove directly that this last expression equals
1 mod $p$, but we can avoid that as follows.
Note that $k \pmod{p}$ is a function of $|G|$ and $p^\beta$ only,
and so it is the same for all groups of order $p^{\alpha} t$. So $k
\pmod{p}$ can be
determined from $G=C_{p^{\alpha} t}$, the cyclic group of order $p^{\alpha} t$.
Hence  $k \equiv 1  \pmod{p}$, since a cyclic group $G$ has a unique
subgroup of each order dividing $|G|$.
$\mathbf{Additional\ Remark\!:}$ This last trick of using the fact that the number of subgroups modulo $p$ depends only on $|G|$ is due to Graham Higman. In his original proof, Wielandt proved directly that the expression in the final displayed formula is congruent to $1$ modulo $p$.
