I've been checking the demonstration of this book (page 34, theorem 1.68) about the construction of an outer automorphism of $S_6$ using Sylow theorems. It's in spanish because that's my first language, but I'm going to summarize the key points.
Let $X$ the set of $5$-Sylow subgroups of $S_5$. Using the Sylow theorems it can be shown that $|X|=6$. Since $p$-Sylow subgroups are conjugated, $S_5$ acts transitively on $X$. Let $\varphi:S_5\rightarrow S_X\cong S_6$ this action. Then, $\ker \varphi=\cap_{P\in X}N(P)$, where $N(P)$ is the normalizer of $P$ in $S_5$, is a normal subgroup of $S_5$ with index $\geq6$. Hence, $\ker\varphi=\{1\}$, $K=\varphi(S_5)\cong S_5$ and $K$ is a transitive subgroup of $S_6$. Let $H=\{\sigma\in S_6\mid\sigma(6)=6\}\cong S_5$, which is a non transitive subgroup of $S_6$.
We want to show an automorphism $\psi:S_6\rightarrow S_6$ such that $\psi(H)=K$. Now, $\psi$ must be an outer automorphism, because if $\psi$ is an inner automorphism, $K$ and $H$ will be conjugated and since $H$ is non transitive, then $K$ will be non transitive, which is a contradiction.
In order to construct $\psi$, let us consider the action of $G=S_6$ on $G/H$ and $G/K$, the set of left cosets of $H$ and $K$, respectively, by left translation. Let $\rho:G\rightarrow S_{G/H}$ and $\xi:G\rightarrow S_{G/K}$ these actions. It can be shown that $\ker\rho=\{1\}=\ker\xi$, hence $\rho$ and $\xi$ are injections and, therefore, surjections.
Now, let $\chi:S_{G/H}\rightarrow S_{G/K}$ the morphism induced by an arbitrary bijection $\eta:G/H\rightarrow G/K$ such that $\eta(H)=K$. Later, it's claim that exists an unique automorphism $\psi$ such that $\xi\circ\psi=\chi\circ\rho$ and $\psi(H)=K$.
My questions are:
- How $\eta$ induces a morphism $\chi$?
- Why $\psi$ is unique and how satisfies $\psi(H)=K$?
Thank you for your time and effort.