Could you perhaps write you proof based on mine ? That would be more helpful. Thank you for you help

@Carpediem If you are looking at a cycle decomposition that contains only 2-cycles which looks like $(ij)(kl)w$ where $w$ is another permutation, then your original permutation is in the alternating group if and only if $w$ is in there.

So what is wrong with what I wrote ?

Nothing is wrong with it, as long as what you're reducing has a cycle of length $k\geq 3$ out in front. If it doesn't, what do you do? If it has a cycle of length 2 out in front, you have a problem. But you don't if there's a cycle of length 2 after that by my previous comment. What if you have a 2-cycle then a k-cycle with $k\geq 3$? Still have a problem. You need to break up that 2nd cycle into a 2-cycle and another cycle. How do you do that?

Ok, how about this? If you have a 2-cycle and a k-cycle out front, you can conjugate by the inverse of the 2-cycle. This new element will be in $A_n$ if and only if the original was. That will put the k-cycle out front.

Oh now I understand ! But the condition $k \geq 3$ is imposed. In this case, do you think I should modify in any way my proof ?

Oh now I understand ! But the condition $k \geq 3$ is imposed. In this case, do you think I should modify in any way my proof ?

I believe after taking those cases into account, you're fine. It's just a little messy. I think there's a slight worry in that how does your method show that you're not going to accidentally reduce a 2-cycle to the identity.

but I don't have to consider the 2-cycle case.. or do I ?

@Bryan

With your method, you show that you can reduce a k-cycle to a k-2 cycle. Using the conjugation method, you can 'move' 2-cycles to the right. So you can 'reduce' an element of $A_n$ to a product of 2-cycles. And you can cancel any pair of 2-cycles on the left. So... yeah I think you're okay.

Wait

here you go. If you have the cycle $(ij)(kl)$ out front, multiply on the left by $(lkj)(jli)$. This will cancel those 2-cycles. If you have the cycle $(ij)(ik)$ on the left, multiply by $(kji)$. That will cancel that pair of cycles.

@Bryan I forgot just one thing

We have to consider the case when p is expressed as 1-cycles and 2-cycles

Since p is even we do not have to consider the case of transpositions and so

p=(12)(34)

So I multiply again by (321)(12)(34)

and obtain (1)(234)...

More fixed indices again

and we're done