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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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    Please, use math typesetting next time you want to post a question. Feel free to edit the modifications I did (which are purely for display purposes, no text has been changed).2012-04-14
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    @Andre : I am also trying to find a hole in his proof. But your subgroup has order $6$, and $6! = 120$, so $m = 120/6 = 20 > n = 6$ which does not satisfy his criterion. You're not giving a counter example here.2012-04-14
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    @Makoto Kato : There is a page on this website that shows how to do it (in the FAQ I think). It really helps rendering, plus it adds to the clarity/readability of your question.2012-04-14
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    I am not finding a reasonable explanation as to why $N$ is a subgroup of $H$ : all I have is that if $g' \in N$, then for all $g \in G$, $$ g' \cdot (gH) = gH \quad \Rightarrow \quad g' gh = gh' $$ for some $h,h' \in H$. But that means $g' = gh'hg^{-1}$, which unless $H$ is normal in $G$, does not mean $g' \in H$. I think it needs clarification.2012-04-14
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    @MakotoKato, your proof is correct. The technique is relatively commonly used. Another result in the same vein that is often mentioned in this site is the following: If $H$ is a subgroup of $G$ of index $p$, where $p$ is the smallest prime dividing $|G|$, then $H$ must be normal. The proof follows quickly from a study of the homomorphism $f:G\to Sym(G/H)$ in that case as well. So, yes, the result is well known, but you did well in figuring it out on your own!2012-04-14
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    @Patrick, the fact that $N=\ker f$ is a subgroup of $H$ follows immediately from the fact that $H$ is the stabilizer of the coset $H$, and $N$ is the subgroup that stabilizes all the cosets. IOW $N$ is the intersection of all the conjugates of $H$.2012-04-14
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    @MakotoKato: Just type a dollar sign `$` before and after each symbol or formula, underline `_` for subscripts (which you did already) and hat `^` for exponents, which you enclose in braces `{` `}` if they are more than one character long, and you'll be on your way to typesetting math (regardless of our age ;-).2012-04-14
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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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