Wielandt's proof of Sylow's theorem.

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 ${\rm Stab} _G(S) \subseteq S$.

1) 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 $

2) 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 $|G|$.