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I have come across this problem in the course of studying for my comprehensive exam in algebra, which is fast approaching. I would appreciate any suggestions and/or hints as to its solution.

Let $S$ be an infinite set, and let $G:=\mathrm{Sym}(S)$, the symmetric group on $S$. Consider the subgroup $A$ of $G$ given by $A:=\left\langle\left(a\hspace{2pt}b\right)\left(c\hspace{2pt}d\right)\mid a,b,c,d\in S\right\rangle$, where the notation $\left(x\hspace{2pt}y\right)$ denotes the transposition of the elements $x$ and $y$. Prove that $A$ is a simple group.

The subgroup $A$ is, essentially, the analogue of the alternating group for finite sets $S$. However, I have as yet been unable to figure out how to apply the proofs of the simplicity of $A_n$ for $n\geq 5$ to this problem (if that is even a valid strategy). Any suggestions would be appreciated.

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  • $\begingroup$ I'm guessing that $\,A\,$ is generated by the product of those two transpositions as $\,a,b,c,d\,$ run over all the quartets (foursomes) of elements of $\,S\,$ ...? $\endgroup$ – DonAntonio Aug 14 '13 at 0:22
  • $\begingroup$ Yes. $A$ is generated by products of an even number of transpositions. $\endgroup$ – anonymous Aug 14 '13 at 0:23
  • $\begingroup$ $a\ne b$ and $c\ne d$, but otherwise, there are no restrictions. $\endgroup$ – anonymous Aug 14 '13 at 0:24
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I think that you can just use the fact that the finite alternating groups $A_n$ are simple for $n\geq5$, without going back to the proof of that fact. First show (if you haven't already done so) that a group is simple iff, for every non-identity element $a$, the conjugates of $a$ generate the whole group. Then, knowing that copies of finite $A_n$'s inside your $A$ have this property, it should be easy to prove the same property for $A$ itself.

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