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Each permutation in $A_k$ can be written as a product of 3-cycles of the form (1, 2, 3), (1, 2, 4),...,(1, 2, k). I am trying to start this problem by induction but I am having trouble with the base case... I tried for k=3, (123)=$(123)^3$, (312)=$(123)^5$, (321)=(123)(123), (231)=(123), but I can't figure out (213) and (132).

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    $\begingroup$ But you already did, it is $(321)$. $\endgroup$ – André Nicolas Mar 17 '16 at 2:20
  • $\begingroup$ I have to write it using (123) per the problem... I think $\endgroup$ – MathIsHard Mar 17 '16 at 2:27
  • $\begingroup$ It is $(123)(123)$, since $(321)$, $(213)$, and $(132)$ are the same permutation. $\endgroup$ – André Nicolas Mar 17 '16 at 2:33
  • $\begingroup$ Thanks so much. I appreciate it. $\endgroup$ – MathIsHard Mar 17 '16 at 2:41
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In $A_3$ the three permutations are $(1)=(123)(123)(123), (123), (132)=(123)(123)$. Not sure where your confusion lies.

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  • $\begingroup$ aren't there 6 permutations? Sorry I am learning this... Also I have to write them using (123) $\endgroup$ – MathIsHard Mar 17 '16 at 2:28
  • $\begingroup$ No need to apologize. Indeed $A_3$ has 3 permutations. You may be confusing $A_3$ with $S_3$ which has 6 permutations. $\endgroup$ – Shahab Mar 17 '16 at 2:29
  • $\begingroup$ Oh I am confusing the two... What is A then and what is S? $\endgroup$ – MathIsHard Mar 17 '16 at 2:36
  • $\begingroup$ $A_3$ is a subgroup of $S_3$ consisting of precisely the even permutations. You may ask a separate question in this regard. $\endgroup$ – Shahab Mar 17 '16 at 2:38
  • $\begingroup$ Oh I had no idea. thank you. I think I understand even permutations enough to work my way through it. $\endgroup$ – MathIsHard Mar 17 '16 at 2:40

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