When I read an article about Cauchy's theorem (1815) on a permutation group, I tried to prove it and I came up with the following proposition which is similar to the Cauchy's but is more general. Is this well-known? If yes, where can I find the proof in an existing literature?
Proposition
Let $n$ be an integer greater than $4$.
Let $S_n$ be the symmetric group of degree $n$.
Let $H$ be a subgroup of $S_n$.
Let $m$ be the number of left cosets of $H$ in $S_n$, i.e. $m = (S_n : H)$, the index of $H$ in $S_n$.
If 1 < m < n, then $m$ must be $2$.
My proof:
Let $G = S_n$.
Let $A_n$ be the alternating group of degree $n$.
It is well-known that $A_n$ is a simple group.
It is an easy consequence of this fact that the only non-trivial normal subgroup of $G$ is $A_n$.
Let $G/H$ be the set of left cosets of $H$.
Let $\mathrm{Sym}(G/H)$ be the symmetric group on $G/H$.
Let $f:G → \mathrm{Sym}(G/H)$ be the homomorphism induced by the left actions of $G$ on $G/H$.
Let $N$ be the kernel of $f$. $N$ is a subgroup of $H$.
Since $|G| = n!$ and $|\mathrm{Sym}(G/H)| = m!$, $f$ can't be injective.
Since the only non-trivial normal subgroup of $G$ is $A_n$, $N = A_n$.
Since $(G : N) = 2$, $H = N$. Hence $m = 2$. Q.E.D.