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.

  • $\begingroup$ 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
  • $\begingroup$ Note: the resulting group is called the alternating group on the set $S$. It is the unique minimal normal subgroup of the group of permutations of $S$, and has index 2 in the group of finitely supported permutations of $S$. $\endgroup$ – YCor Jul 15 '19 at 14:46

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.