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Let $H$ be a subgroup of $G$ with $[G:H]=n$ and $G$ finite. Let $G$ act on left cosets of $H$ by left multiplication. Let $N$ be the kernel of the action. Show that $[G:N]$ divides $n!$.

Ok so I have a theorem that says that if $G$ is simple then it embeds into $S_n$. So if $G$ is simple than I'm done because $G$ divides $n!$ and thus $N$ divides $n!$ and hence $[G:N]$ divides $n!$.

Now I also know that $N$ is the largest normal subgroup of $G$ contained in $H$. Thus if $G$ isn't simple then it contains some non-trivial normal subgroup $K$ and thus $K\cap H$ is normal in $H$ and by the 2nd Iso Theorem $H/K\cap H\cong KH/H$...

As you can see I'm not sure where to go from here. I was hoping to prove that $K\cap H$ wasn't trivial and thus that $N$ couldn't be trivial since it must contain $K\cap H$, but I'm not sure I can, and even if I could I'm not sure that would help me in showing that $[G:N]$ is small enough to divide $n!$.

Can anyone give some guidance? Thanks.

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Ronald Weasley! –  KUSH Nov 28 '12 at 0:32

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up vote 3 down vote accepted

Let $S_{G:H}$ be the set of permutations of $G:H$. Consider $\phi:(G:N)\rightarrow S_{G:H}$ that sends $xN$ to $T_x$ (Where $T_x$ is the permutation of $G:H$ that sends $aH$ to $xaH$ ). It is easy to verify that $\phi$ is an injective group homomorphism. Thus, $G:N$ is isomorphic to a subgrop of $S_{G:H}$. Using Lagrange's theorem, we deduce that $|G:N|$ divides $|S_{G:H}|=n!$

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I'm assuming by isomorphism you mean embedding (injective hom). I've been completely unsuccessful proving that $\phi$ is an embedding. For instance for injectivity: I assume $gaH=faH$ for all $a\in G$ and I need to prove that $gN = fN$, but the most I can prove is that $gf^{-1}\in H$ when what I need to show is that $gf^{-1}\in N$. –  cactuar Nov 26 '12 at 1:02
    
OH yes. I will edit it –  Amr Nov 26 '12 at 1:05
    
any ideas for proving injectivity? –  cactuar Nov 26 '12 at 1:24
    
Let $\phi(x_1)=\phi(x_2)$, therefore $T_{x_1}=T_{x_2}$. Hence for all a in G $x_1aH=x_2aH$. Thus $x^{-1}_2x_1aH=aH$ therefore $x^{-1}_2x_1\in N$. Hence $x_1N=x_2N$ –  Amr Nov 26 '12 at 1:37
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You should define $\phi$ on $G$ and use the homomorphism theorem in order to avoid redundant computations. –  Martin Brandenburg Nov 26 '12 at 1:52

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