# How to show that the 3-cycles $(2n-1,2n,2n+1)$ generate the alternating group $A_{2n+1}$.

I'm asked to show that the 3-cycles $(1,2,3),(3,4,5),(5,6,7),...$ and $(2n-1,2n,2n+1)$ generate the alternating group $A_{2n+1}$.

I know the 3-cycles produce the group $A_n$, and it seems like I have to use that. Furthermore I know that $(123)$ and the $k$-cycle $(123...k)$ generates $A_k$ with $k$ odd. Seeing that the group $A_{2n+1}$ is odd this seems like a good way to tackle the problem, but from here on I'm stuck.

This is not the only question I'm unable to answer about permutation groups. In general I can't visualize them or see any obvious pattern in their behavior.

Any help would be appreciated.

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Since the product of $(1,2,3)$ and $(3,4,5)$ is a 5-cycle, you can generate $A_5$ on $\{1,2,3,4,5\}$. So you have all 3-cycles of the form $(a,b,5)$ with $a,b \in \{1,2,3,4\}$, and they enable you to generate $A_5$ on $\{a,b,5,6,7\}$, so you have all 3-cycles on $\{1,2,3,4,5,6,7\}$, etc. etc. –  Derek Holt Jan 5 '13 at 21:42

Try approaching this via induction on $n$?
You know that $(1,2,3)$ generates $A_3$ and $(1, 2, 3)(3,4,5)$ is a $5$-cycle, so you can generate $A_5$ on $\{1,2,3,4,5\}$. So you have $3$-cycles of the form $(a,b,5)$ with $a,b\in\{1,2,3,4\}$, and that lets you generate $A_5$. Then note that in $\{a,b,5,6,7\}$ you can find the needed 3-cycles on $\{1,2,3,4,5,6,7\}$ that generate $A_7$...
Assume this holds in terms of the three-cycles generating $A_{2n - 1}$, then show it holds for $A_{2(n+1)-1} = A_{2n+1}.$