Prove that the alternating group $A_{n}$ is generated by all products of two 2-cycles ($i$,$j$)($k$,$l$). These 2-cycles are not necessarily disjoint.

So I know that $A_{n}$ can be written as products of even number of transpositions. In order to prove the statement, we have to prove that we can write any element in $A_{n}$ as ($i$,$j$)($k$,$l$). Equivalently, we have to prove that we can write any transposition using any product of two 2-cycles ($i$,$j$)($k$,$l$). Is my reasoning correct?

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    $\begingroup$ It is easier than you think. Also note that a transposition is an odd permutation, so there are no transpositions in $A_n$. But if you have a product of an even number of transpositions, how can you see it also as a product of the kind you want? $\endgroup$ – Mark Bennet Nov 7 '18 at 16:48
  • $\begingroup$ @MarkBennet When you say that it is easier than I think, I'm sure that there's something important I have missed. Wondering if you can give me more hint? $\endgroup$ – JJW22 Nov 7 '18 at 16:57

I think you have answered the question yourself. But your second sentence is wrong: "In order to prove the statement, we have to prove that we can write any element in $A_n$ as $(i,j)(k,l)$". That couldn't be true if $n$ were large. When we say that a group is generated by a set of elements, it doesn't mean that every element in the group is one of those; it means that every element can be written as a a finite number of products of them.

  • $\begingroup$ Hi, wondering if you can elaborate on how I have answered my question. Still struggling to see the point here. Thank you! $\endgroup$ – JJW22 Nov 7 '18 at 16:59
  • $\begingroup$ Well you said that every element of $A_n$ is the product of an even number of transpositions. For any such expression, takes the first two transpositions, then the next two, and so on, and you have your proof. $\endgroup$ – Richard Martin Nov 7 '18 at 17:07
  • $\begingroup$ I think I misunderstood the question. I thought that by products of two 2-cycles, it means that we have to prove that every element in $A_{n}$ can be written as a single (i,j)(k,l). $\endgroup$ – JJW22 Nov 7 '18 at 17:10
  • $\begingroup$ Yes, hence my first comment/answer. You wouldn't be able to prove that in, for example, $A_5$, because a single $(i,j)(k,l)$ wouldn't be able to 'move enough things around', e.g. $(12345)$ could not be thus written. $\endgroup$ – Richard Martin Nov 7 '18 at 17:12
  • $\begingroup$ Thank you! Really appreciate your patience! $\endgroup$ – JJW22 Nov 7 '18 at 17:14

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