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I am trying to show that $A_n$ is generated by the 3-cycles in $S_n$. It seems that every 3-cycle of the form $(a_1,a_2,a_3)$ can just be written as $(a_1,a_3)(a_1,a_2)$ so every 3-cycle turns into an even number of transpositions (2-cycles). Is this sufficient?

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    $\begingroup$ Nope, this is not sufficient. What you are showing is that $3$-cycles lie in $A_n$, not that they generate $A_n$. $\endgroup$ – darij grinberg Oct 19 '15 at 3:34
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    $\begingroup$ Try writing $(a,b,c,d,e)$ as a product of $3$-cycles, for example. $\endgroup$ – Akiva Weinberger Oct 19 '15 at 3:37
  • $\begingroup$ (a,b,c,d,e) = (a,d,e)(a,b,c)? $\endgroup$ – letsmakemuffinstogether Oct 19 '15 at 4:14
  • $\begingroup$ In general $(a_1,a_2,\ldots,a_m)=(a_1,\ldots,a_k)(a_k,a_{k+1},\ldots,a_m)$. Iterating this lets you reduce a cycle to a product of transpositions. Knowing this makes it easier to use my answer to solve the problem. $\endgroup$ – Matt Samuel Oct 19 '15 at 4:28
  • $\begingroup$ I see, so perhaps the answer that Akiva Weinberger was looking for is that $(a,b,c,d,e) = (a,b,c)(c,d,e)$. Yes? $\endgroup$ – letsmakemuffinstogether Oct 19 '15 at 4:31
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An element of $A_n$ is a product of an even number of transpositions. For any pair of transpositions, find one or two 3-cycles whose product is equal to their product. After doing this you can generate all products of an even number of transpositions.

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