The proof that this question concerns is from the Sylow's theorems.
Let $G$ be a finite group of order $p^rm$ where $p$ is prime, $r\geq1$ and $p\nmid m$. Proof the existence of at least one subgroup $P$ of order $p^r$.
I am stuck at the part where you show that there exists only one orbit $\{\alpha \}$ of length $1$.
Let $P\leq G$ a maximal subgroup of $G$ with $|P|=p^s$., $H:=N(P)$ the normalizer of $P$. Let $\Omega:= cos(G:H)$ and $\alpha:=H$ so that $H=Stab(\alpha)$. Consider the action of the $P$ on $\Omega$. As $P\leq H$ there is one $P$-Orbit of length $1$ which is $\{\alpha\}$. Be $\{\beta\}$ any $P$- Orbit of length $1$.
Then $\beta=x^{-1}\alpha x$ for some $x\in G$.
What is the reason for this last statement?
I know that if $G$ operates on its subgroups the conjugation is a transitive operation and thus it would be clear. Cosets aren't subgroups though. So I thought it probably has to do with how conjugation is an inner automorphism, thus bijective and $|\{\beta\}|=\{\alpha\}|=1$ but I can't quite put it together.
Are orbits of the same length always conjugate to each other? Is it enough to just say that you can map one coset to another through a bijection? Or can I somehow proof that the operation of $G$ operates on $\Omega$ through conjugation is transitive? I get stuck because I end up with trying to show that there exists a $g\in G$ so that the RHS is true:
$$ aH \sim bH \Leftrightarrow g(aH)g^{-1} = bH $$
(where $a,b\in G$) I don't know how I would show that both cosets are the same.
Relevant part of the proof from Groups and Geometry (Neumann, Stoy, Thompson)
