I'm trying to prove that $A_5$ is perfect. The only proof I found until now is: "It follows from the fact that it's simple and non-abelian".
Simplicity is quite stronger, and since I only need perfectness (perfection?) I wanted to prove it directly. I did however not go very far, but here is what I had:
I want to show that every element $\rho$ of $A_5$ is in $[A_5, A_5] = \langle \sigma \tau \sigma^{-1} \tau^{-1} | \sigma,\tau \in A_5\rangle$. If I can show it for product of two transposition $(a b)(c d)$, then by induction on the (even) number of transposition of $\rho$, I am done. I considered the case $\rho = (12)(34)$, and $\sigma = (ab)(cd), \tau = (ef)(gh)$ but I got nowhere. I don't really see what logic I can apply to find these $\sigma, \tau$: It seems to me that it is some kind of "trick-proof", where it is very hard to find those elements, but once you have them it is trivial to check that it holds.
Any help would be appreciated.
Thanks, Olivier
EDIT: Thanks to everyone. I finally was able to prove it like this: if $(xyz)$ is an arbitrary 3-cycle, and $r,s$ are the 2 other values, then
$$ [A_5,A_5] \ni (xrz)(zys)(xrz)^{-1}(zys)^{-1} = (xrz)(zys)(xzr)(zsy) = (xrzys)(xzsyr)=(xyz) $$
Hence every 3-cycle is a commutator, and since each $\sigma \in A_5$ is a product of 3-cycles, we are done. I have only one question left: I found this by trying randomly commutators of 3-cycles, and I don't feel very smart for it. Was there a better way?
Show that every 3-cycle is a commutator.