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 $G$.