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July 5th, 2017, 04:22 PM  #1 
Newbie Joined: Feb 2017 From: Argentina Posts: 2 Thanks: 0  Every element of A_5 is a commutator (group theory).
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+m31. Only problem is the three images may not be disjoint. But because A_5 is simple, perhaps this is necessarily so. Do I go somewhere with this line of reasoning? 
July 6th, 2017, 10:07 AM  #2 
Member Joined: Jan 2016 From: Athens, OH Posts: 93 Thanks: 48 
I definitely approve of your usage of exponentiation for conjugation. However, by inference from your post, you write $x^g=gxg^{1}$. Every group theory book that I know (doesn't mean there are not books with your definition) defines $x^g=g^{1}xg$. For example, this way an "obvious" law $x^{gh}=(x^g)^h$ is true. Also the commutator is usually defined as $[a,\,b]=a^{1}b^{1}ab$. You have correctly observed that any nonidentity element of $A_5$ is either a 5cycle, a 3cycle or the product of two disjoint 2cycles. In particular then the order of any nonidentity element of $A_5$ is 5, 3 or 2. Let x be a 5cycle. By Sylow and the simplicity of $A_5$, the normalizer of <x> has order 10. So there is $g\in A_5\text{ of order 2 with }x^g\in <x>$. Hence $x^{1}x^g=[x,\,g]\in <x>$. If $[x,\,g]$ were the identity, then x and g would commute, forcing the order of xg to be 10, but no such element of order 10 exists. So $[x,\,g]$ is a nonidentity element of <x> and so is a 5cycle. Similarly, there is a 3cycle which is a commutator and also there is a product of two 2cycles which is a commutator. I leave the proof to you. If you have trouble, post again and I'll try to help. Finally if $y=[a,\,b]$ is a commutator, then $y^g=[a^g,\,b^g]$ is a commutator for any g. Since any two elements with the same cyclic decomposition are conjugate in $S_5$ and $A_5$ is normal, the above two paragraphs show every element of $A_5$ is a commutator. 

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commutator, element, group, theory 
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