Let \(H\) be a subgroup of \(S_n\) of index \(n\). Prove:
There is an isomorphism \(f: S_n \to S_n\) such that \(f(H)\) is the subgroup of \(S_n\) stabilizing \(n\). In particular, \(H\) is isomorphic to \(S_{n-1}\).
The only subgroups of \(S_n\) containing \(H\) are \(S_n\) and \(H\).
Solution
For \(n=1\), both \(S_1\) and \(H\) are trivial. The identity automorphism proves both assertions, with \(S_0\) also the trivial group. Henceforth suppose \(n\geq2\).
1. The left-coset action defines an injective homomorphism\(\rho:S_n\longrightarrow\operatorname{Sym}(S_n/H)\).
Left multiplication is well defined: replacing a representative \(x\) by \(xh\) with \(h\in H\) does not change \(gxH\). Multiplication by \(g^{-1}\) is its inverse, and the multiplication law in \(S_n\) verifies the action law. Its kernel \(N\) is normal in \(S_n\) and contained in \(H\), since an element fixing all cosets fixes the coset \(H\). In particular \(|N|\) divides \(|H|=(n-1)!\).
For \(n\geq5\), every nontrivial normal subgroup of \(S_n\) contains \(A_n\) [@DF04]. Since \(|A_n|=n!/2>(n-1)!=|H|\), this forces \(N=1\). For \(n=4\), the nontrivial normal subgroups are \(V_4,A_4,S_4\), of orders \(4,12,24\) [@DF04]. None has order dividing \(|H|=6\), so again \(N=1\). For \(n=3\), the only nontrivial proper normal subgroup is \(A_3\), of order three [@DF04]; neither it nor \(S_3\) can lie in the order-two group \(H\). Finally, for \(n=2\) the group \(H\) itself is trivial. Thus the action is faithful in every case.
2. Relabeling the cosets produces the requiredautomorphism \(f\) and the isomorphism \(H\cong S_{n-1}\).
The set \(S_n/H\) has \(n\) elements. Choose a bijection with \(\{1,\ldots,n\}\) that labels the coset \(H\) by \(n\). Through this bijection, \(\rho\) becomes an injective homomorphism \(f:S_n\to S_n\). Both groups have \(n!\) elements, so \(f\) is an automorphism.
An element \(g\) fixes the coset \(H\) exactly when \(gH=H\), which is equivalent to \(g\in H\). Surjectivity of \(f\) therefore gives \[ f(H)=\{\sigma\in S_n:\sigma(n)=n\}=:H_n. \] Restriction to \(\{1,\ldots,n-1\}\) is an isomorphism \(H_n\to S_{n-1}\): any permutation of those letters extends uniquely by fixing \(n\). This proves part (1).
3. The only subgroups containing \(H\) are \(H\) and \(S_n\).
It suffices to prove this for \(H_n\), since \(f\) bijects subgroups and preserves inclusion. Suppose \(H_n\subsetneq J\leq S_n\) and choose \(g\in J\setminus H_n\). Then \(g(n)=i\ne n\). The group \(H_n\) acts transitively on the other \(n-1\) letters, so the \(J\)-orbit of \(n\) contains all letters: it contains \(i\) and every \(h(i)\) with \(h\in H_n\), as well as \(n\). The stabilizer of \(n\) in \(J\) is exactly \(H_n\), because \(H_n\leq J\) and \(H_n\) is the full stabilizer in \(S_n\). Orbit-stabilizer gives \(|J|=n|H_n|=n!\), hence \(J=S_n\). This proves part (2) and, for \(n\geq2\), maximality of \(H\).