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Prove that the simple group of order $660$ is isomorphic to a subgroup of the alternating group of degree $12$.

I have managed to show that it must be isomorphic to a subgroup of $S_{12}$ (through a group action on the set of Sylow $11$-subgroups). Any suggestions are appreciated!

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Once you have that your group (call it $G$) embeds into $S_{12}$, you can apply the signature morphism $\epsilon:S_{12}\rightarrow\lbrace-1,+1\rbrace$. Its kernel is a normal subgroup of $G$. Since $G$ is simple by hypothesis, it is either $G$ or $1$, and by cardinality reasons it has to be all of $G$. Thus $G$ maps into the kernel $\mathrm{Ker}(\epsilon)=A_{12}$. –  Olivier Bégassat Jul 21 '12 at 2:43
    
@OlivierBégassat, you should write your comment as an answer to upvote it...nice! –  DonAntonio Jul 21 '12 at 3:17
    
@DonAntonio ok, done :) –  Olivier Bégassat Jul 21 '12 at 3:40
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2 Answers 2

up vote 15 down vote accepted

This is a comment made into an answer.

Using the Sylow theorems, if $G$ is a simple group of order $660=60\times 11$, the number of Sylow-$11$ subgroups $n_{11}$ has $n_{11}\equiv 1~\mathrm{mod}~11$, $n_{11}|60$ and $n_{11}\neq1$ by simplicity, so $n_{11}=12$. Again, by Sylow's theorem, $G$ acts transitively on the $12$-element-set of its $11$-Sylows by conjugation, and so we get a non trivial morphism $G\rightarrow S_{12}$. By simplicity of $G$, this is an embedding.

Now apply the signature morphism $\epsilon:S_{12}\rightarrow\lbrace-1,+1\rbrace$: $$G\hookrightarrow S_{12}\rightarrow \lbrace-1,+1\rbrace$$ The kernel is a normal subgroup of $G$. Since $G$ is simple, it is either $G$ or $1$, and by cardinality reasons it has to be $G$. Thus the embedding $G\hookrightarrow S_{12}$ maps $G$ into $\mathrm{Ker}(\epsilon)=A_{12}$, and $G$ is isomorphic to a subgroup of $A_{12}$.

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+1 Nicely done. –  DonAntonio Jul 21 '12 at 11:55
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Let $G$ be our simple group of order $660$, contained in $S_{12}$. Since $A_{12}$ is normal in $S_{12}$, the subgroup $A_{12} \cap G$ is normal in $G$. Since $G$ is simple and any subgroup of $S_{12}$ with more than two elements contains an even permutation, we get $A_{12} \cap G = G$.

To generalize, suppose $G$ is a simple group with $|G| > 2$. With the same idea as before, you can show that if $G$ is isomorphic to a subgroup $H$ of $S_n$, then $H$ is a subgroup of $A_n$.

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