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

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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+m3-1. 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 non-identity element of $A_5$ is either a 5-cycle, a 3-cycle or the product of two disjoint 2-cycles. In particular then the order of any non-identity element of $A_5$ is 5, 3 or 2. Let x be a 5-cycle. By Sylow and the simplicity of $A_5$, the normalizer of has order 10. So there is $g\in A_5\text{ of order 2 with }x^g\in$. Hence $x^{-1}x^g=[x,\,g]\in$. 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 non-identity element of and so is a 5-cycle. Similarly, there is a 3-cycle which is a commutator and also there is a product of two 2-cycles 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. Tags commutator, element, group, theory Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post msv Abstract Algebra 1 February 19th, 2015 12:19 PM cummings123 Abstract Algebra 1 February 3rd, 2013 11:23 AM Watari Abstract Algebra 2 December 13th, 2012 06:59 PM butabi Abstract Algebra 8 September 3rd, 2011 02:52 PM DLowry Abstract Algebra 2 April 25th, 2011 11:43 PM

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