Set of generators of the alternating group $A_9$ . The group $A_9$ is generated by $(14)(29)(37)(56)$ and $(123)(456)(789)$.

Question

I was reading K. Conrad's paper on the group $SL_2(\mathbb{Z})$. There he says the group $A_9$ turns out to be generated by $(14)(29)(37)(56)$ and $(123)(456)(789)$. I want to show this.

My attempt: We know that alternating group $A_n$ is generated by all $3$ cycles. In fact I can xeduce that all $3$ cycles of the form $(12j)$ where $j$ is an integer such that $2<j\le n$ generate $A_n$. Now I want to show any $3$-cycle in $A_9$ of the form $(1 2 j)$ where $j$ in $\{ 3,4,5,6,7,8,9\}$ can be expressed as a product of $(123)(456)(789)$ and $(14)(29)(37)(56)$.

I'm stuck on how to show this.

Answer 1

Let your generators be $a$ and $b$ and $G = \langle a,b \rangle$.

Then $ab$ is a product of a $3$-cycle and a $5$-cycle, so $G$ contains $3$-cycles.

Also $abab^{-1}$ is a $7$-cycle. Now $G$ is transitive, and a point stabilizer contains a $7$-cycle and $a$, which is a product of four $2$-cycles, so the point stabilizer is transitive on the remaining points, hence $G$ is $2$-transitive, and the $7$-cycle means that it is $3$-transitive.

So $G$ contains all $3$-cycles, and hence $G=A_9$.