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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.

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    @Makato: a proof of this theorem that doesn't invoke simplicity of $A_n$, but is an argument similar to the one you give, is in Theorem 2.7 of http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/Ansimple.pdf.2012-04-15

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