Every element of A_5 is a commutator (group theory).

[STF92]

A_5 is the alternate group of degree 5. I found a proof on Internet. It just gives (ijk), (ij)(kl), (ijklm) as certain commutators. But how did he found them? There are three conjugate classes in A_5. So x^g, the conjugate of x by g sweeps its class as x sweeps that class. So x^1x^g also does. But x^1x^g is [x,g]. So if the class has size m1, then I get m1 commutators. Doing this for each of the three classes I get m1+m2+m3 commutators. But |G|= m1+m2+m3-1. Only problem is the three images may not be disjoint. But because A_5 is simple, perhaps this is necessarily so.

 

[stapel]

A_5 is the alternate group of degree 5. I found a proof on Internet. It just gives (ijk), (ij)(kl), (ijklm) as certain commutators. But how did he found them? There are three conjugate classes in A_5. So x^g, the conjugate of x by g sweeps its class as x sweeps that class. So x^1x^g also does. But x^1x^g is [x,g]. So if the class has size m1, then I get m1 commutators. Doing this for each of the three classes I get m1+m2+m3 commutators. But |G|= m1+m2+m3-1. Only problem is the three images may not be disjoint. But because A_5 is simple, perhaps this is necessarily so.

Who is he? Where did he post... whatever he posted?

 

[STF92]

Well, rather it is a PDF file I found on the Internet, dated march 9, 2009. It seems some excerpt from a book. The proof is this:

Lemma 120. Every element of the alternating group A5 is a commutator.

Proof. Every element of A5 is of the form (i jk), ( i j)(k l) or (i jklm) where
i, j, k, l , m ∈ {1, 2, 3, 4, 5 } are distinct. The following calculations finish the
proof:
[( i jl) , (ikm )] = ( i jl)(ikm)(il j)(imk) = ( i jk ) ,
[( i jk ) , (i jl)] = (i jk)(i jl)(ik j)(i l j) = (i j)(kl),
[( i j)(km), ( iml)] = (i j)(km)(iml)(i j)km (ilm) = (i jklm).