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Prove that, for $n \geq 3$, the three-cycles generate the alternation group $A_n$

Proof: We multiply on the left by 3-cycles to "reduce" an even permutation $p$ to the identity, using induction on the number of indices fixed by a permutation. How the indices are numbered is irrelevant. If $p$ contains a $k$-cycle with $k \geq 3$, we may assume that it has the form $p=(123\dots k)\dots$ Multiplying on the left by $(321)$ gives $$p'= (321)(123 \dots k)\dots=(1)(2)(3\dots k)\dots$$ More fixed indices.

What do you think ?

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  • $\begingroup$ How would you use your method to show that for instance $(12)(34)$ may be written as a product of 3-cycles? $\endgroup$ – paw88789 Aug 30 '14 at 22:39
  • $\begingroup$ Hi Carpe, But I thought you were multiplying by 3-cycles to get to the identity. $\endgroup$ – paw88789 Aug 30 '14 at 22:45
  • $\begingroup$ mmm.. Let me think about that. $\endgroup$ – Carpediem Aug 30 '14 at 22:48
  • $\begingroup$ Where have you used he fact that the permutation $p$ is even? If your argument worked, then why would it not "prove" that $S_{n}$ is generated by $3$-cycles (which it isn't)? $\endgroup$ – Geoff Robinson Aug 30 '14 at 23:43
  • $\begingroup$ Technical point: it is better to use Markdown syntax for **bold** font, because then the font matches the rest of text, unlike the font used by $\textrm{MathJax}$. $\endgroup$ – user147263 Aug 31 '14 at 4:29
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A permutation is an element of $A_n$ if and only if it is a product of an even number of transpositions.

We first note that the three cycles do not generate more than $A_n$ since for distinct $i,j,$ and $k$, we have $(ij)(ik)=(ijk)$.

For the other inclusion, we note that $(ijk)=(ij)(ik)$ for distinct $i,j,$ and $k$. And for distinct $i,j,k,$ and $l$, we have that $(ij)(kl)=(ijk)(jkl)$. Thus the three cycles generate all products of an even number of transpositions.

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  • $\begingroup$ Could you perhaps write you proof based on mine ? That would be more helpful. Thank you for you help $\endgroup$ – Carpediem Aug 30 '14 at 22:57
  • $\begingroup$ @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. $\endgroup$ – Robert Wolfe Aug 30 '14 at 23:00
  • $\begingroup$ So what is wrong with what I wrote ? $\endgroup$ – Carpediem Aug 30 '14 at 23:04
  • $\begingroup$ 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? $\endgroup$ – Robert Wolfe Aug 30 '14 at 23:09
  • $\begingroup$ 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. $\endgroup$ – Robert Wolfe Aug 30 '14 at 23:11
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The idea of the proof is that we will show that any product of two transpositions is a product of 3-cycles ; and ,if so, then since every member of the An is a product of even number of transpositions, we are done.

To begin with lets us suppose that we have a product of transpositions a1 and a2 where both have a common number a which belongs to the set {1,2,.....}- say they both move a. Then we have a1 of the form (a b) and a2 of the form (a c) . In this case , we have

a1 a2 = (a b)(a,c)= (a,c,b) and we are done. Now suppose that a1 and a2 move different numbers .

a1 = (a b) a2 = (c d) Then

a1a2 = (a,b)(c,d) = ( d a c ) ( a b d ). And we are done.

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