# Permutation and induction

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).

• But you already did, it is $(321)$. – André Nicolas Mar 17 '16 at 2:20
• I have to write it using (123) per the problem... I think – MathIsHard Mar 17 '16 at 2:27
• It is $(123)(123)$, since $(321)$, $(213)$, and $(132)$ are the same permutation. – André Nicolas Mar 17 '16 at 2:33
• Thanks so much. I appreciate it. – MathIsHard Mar 17 '16 at 2:41

## 1 Answer

In $A_3$ the three permutations are $(1)=(123)(123)(123), (123), (132)=(123)(123)$. Not sure where your confusion lies.

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