I'm reading a proof that every subgroup of $S_n$ with index n is isomorphic to $S_{n-1}$. I have some questions about the proof:
Let $H$ be a subgroup of $S_n$ with index $n$. Consider the left multiplication action of $S_n$ on $S_n/H$ which is the morphism $l:S_N\rightarrow Sym(S_n/H)$. Since $S_n/H$ has a size $n$, $Sym(S_n/H)$ is isomorphic to $S_n$. The kernel of $l$ is a normal subgroup of $S_n$ which lies in $H$. Thus, the kernel of has index at least $n$ in $S_n$. Since the only normal subgroups of $S_n$ are identity, $A_n$, $S_n$, the kernel is trivial hence $l$ is an isomorphism. What is the image of $l(H)$ in $Sym(S_n/H)?$ Since $gH = H \Leftrightarrow g\in H$, $l(H)$ is the group of permutations of $S_n/H$ which fixes the point $H$ in $S_n/H$. The subgroup fixing a point in a symmetric group isomorphic to $S_n$ is isomorphic to $S_{n-1}.$
(1) When it says since $gH = H$ if and only if $g\in H$, $l(H)$ is the group of permutation of $S_n/H$ which fixes the point $H$ in $S_n/H$. I don't understand how this conclusion was reached?
(2) The subgroup fixing a point in a symmetric group isomorphic to $S_n$ is isomorphic to $S_{n-1}$. How is this statement proved? Is the subgroup mentioned, the stabiliser of $S_n$ for which the proof is pretty simple and can be found on the stack?
Also, a sketch of a solution to the theorem is given at Subgroups of $S_n$ of index $n$ are isomorphic to $ S_{n-1}$. At the end as the OP claims the idea is to show that the homomorphism $\phi: G \rightarrow Aut(S_n/G \backslash \{G\})\cong S_{n-1}$ is isomorphic if it's shown to be injective. How exactly does this argument work out? Also, Why is it claimed that the $\phi \cong S_{n-1}$, isn't proving this the whole purpose of the problem?
Best Answer
An element $g\in S_n$ has $\ell(g)\in \{\sigma\in \mathrm{Sym}(G/H)\mid \sigma(H)=H\}$ if and only if $\ell(g)H = H$, if and only if $gH=H$, if and only if $g\in H$. So the elements of $G$ that map to the stabilizer of $H$ in $\mathrm{Sym}(G/H)$ under $\ell$ are precisely the elements of $H$. But since $\ell$ is an isomorphism, that means that $\ell(H)$ is the point stabilizer of $H$ in $\mathrm{Sym}(G/H)$.
What are the permutations of $S_n$ fixing, say, $n$? $\sigma\in S_n$ fixes $n$ if and only if it sends the subset $\{1,\ldots,n-1\}$ to itself, and so the restriction of the point stabilizer of $n$ to $\{1,\ldots,n-1\}$ is isomorphic to $S_{n-1}$. Now note that if $M$ is the stabilizer of $n$ in $S_n$, and $\tau=(i,n)$, then $\tau M\tau$ is the stabilizer of $i$; since $\sigma\mapsto\tau\sigma\tau^{-1}=\tau\sigma\tau$ is an automorphism, the stabilizer of $i$ is isomorphic to $M$ is isomorphic to $S_{n-1}$.
Your final paragraph is garbled. The fact that $\mathrm{Sym}(S_n/G\setminus\{G\})$ is isomorphic to $S_{n-1}$ is simply that the set $S_n/G\setminus\{G\}$ has exactly $[S_n:G]-1=n-1$ elements, and if $|X|=|Y|$, then $\mathrm{Sym}(X)\cong\mathrm{Sym}(Y)$. If you can show $\phi$ is an isomorphism, then its domain, $G$, is isomorphic to its codomain, $\mathrm{Sym}(S_n/G\setminus\{G\})$, which is naturally isomorphic to $S_{n-1}$, so this will show $G\cong S_{n-1}$. As to “how it will go”... it will go exactly the same way as the proof you are quoting, with $G$ playing the role of $H$. It’s the same proof.