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I was reading a proof that any finite group can be embedded into the alternating group $A_{n}$, and it mentioned that in order to show this, it is sufficient to show that $Sym(n)$ can be embedded into $A_{2n}$.

Could somebody please explain to me why this is?

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    Because any group $G$ can be embedded into $Sym(|G|)$ (the set of bijection on its underlying set). For each $g$ in $G$ use $g \mapsto (x \mapsto gx)$. – Nex Nov 06 '16 at 04:46
  • @Nex, any group $G$ can be embedded into the group of symmetries on however many elements there are in $G$? And that map that you have is a double map? Anyway, why $A_{2n}$? You didn't really answer that. –  Nov 06 '16 at 04:49
  • Surely the original statement says that every finite group can be embedded into $A_n$ for some $n$? Writing $G$ for the group, we embed $G$ into $A_{2|G|}$ by composing the two embeddings $G$ into $S_{|G|}$ and $S_{|G|}$ into $A_{2|G|}$. – Nex Nov 06 '16 at 04:53

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