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Let $G$ be an infinite group , $e\neq x\in G$ and the conjugate class of $x$ has finitely many elements. Prove that $G$ is not a simple group.

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By hypothesis you have that your group $G$ acts on the conjugate class of $x$ which is a finite set, let call $n$ the cardinality of this set. So $G$ acts on this set of $n$ element, this gives us an homomorphism $\varphi \colon G \to S_n$, where $S_n$ is the group of permutation on $n$-elements. Clearly this homomorphism cannot be injective, because $G$ is infinite while $S_n$ has just $n!$ elements, so this homomorphism must have a non trivial kernel which is a normal subgroup of $G$.

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This doesn't cover the case $n=1$. –  Chris Eagle Dec 30 '12 at 13:03
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That case $\,n=1\,$ is trivial, @ChrisEagle, as $\,|Conj(x)|=1\Longleftrightarrow x\in Z(G)\,$, but since $\,x\neq e\,$ then this would mean $\,1\neq Z(G)\triangleleft G\,$ and we're done. –  DonAntonio Dec 30 '12 at 13:13
    
@ChrisEagle You're right, but it seems that DonAntonio beat me in time. Anyway thank you both to helping me improving the answer. :) –  Giorgio Mossa Dec 30 '12 at 13:55

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