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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). –  Patrick Da Silva Apr 14 '12 at 7:21
    
@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. –  Patrick Da Silva Apr 14 '12 at 7:26
    
@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. –  Patrick Da Silva Apr 14 '12 at 7:29
    
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. –  Patrick Da Silva Apr 14 '12 at 7:34
    
@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! –  Jyrki Lahtonen Apr 14 '12 at 7:42

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