The alternating group $A_n$ is generated by $\langle(1 2 \dots n) , (123)\rangle$

Question

Show that the alternating group $A_n$ is generated by $\langle(1 2 \dots n) , (123)\rangle$

(Let $n$ be an odd number.)

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I already know that all $3$-cycle generate $A_n$ but how can I show this other problem?

First, $(1 2 3 \dots n) (321) = (1 4 5 \dots n)$ Then, we can make $(1 2 \dots n-3 \; \; n-2) $ by conjugation of n-cycle ,4 to 1 , 5 to 2 , 6 to 3 , 7 to 3 ...etc.

so, $(1 2 \dots n-3 \; \; n-2)(321)=(1 4 5 \dots n-2)$

so , we can make $(12345)$ by deleting two element again and again.

How can I proceed?

Answer 1

Note that we must assume the $n$ is odd.

Let $H=\langle(1\,2\,3\,\ldots\,n),(1\,2\,3)\rangle$. If $n=3$, then already $H=A_3$. Hence we may assume $n\ge5$. In that situation, you showed that also $(1\,2\,3\,\ldots\,n-2)\in H$. Hence we may assume by induction that $\langle (1\,2\,3\,\ldots\,n-2),(1\,2\,3)\rangle =A_{n-2}\subseteq H$.

Now consider $g\in A_n$, where $n>3$. Then there exist $h\in H$ such that $hg$ maps $n\mapsto n$, for example $h=(1\,2\,\ldots\,n)^{n-g(n)}$ has this property. Among all $h\in H$ with $hg(n)=n$, pick one that maximizes $hg(n-1)$. Assume $hg(n-1)<n-1$. Certainly $hg(n-1)\ge 2$ as otherwise $(1\,2\,3)h$ contradicts maximality of $h$. Consider $$h'=(1\,2\,\ldots\,n)^r(1\,2\,3)(1\,2\,\ldots\,n)^{-r}h=(r+1\,r+2\,r+3)h,$$ where $r=hg(n-1)-2$. Then $$h'g(n) =(r+1\,r+2\,r+3)hg(n)=(r+1\,r+2\,r+3)n=n$$ because $r+3<n$, and $$h'g(n-1)=(r+1\,r+2\,r+3)(r+2)=r+3=hg(n-1)+1,$$ contradicting maximality of $h$. We conclude that $hg(n)=n$ and $hg(n-1)=n-1$, so $hg\in A_{n-2}$ and finally, $g\in h^{-1}A_{n-2}\subseteq H$.

Answer 2

First, you need $n$ to be odd for this to be true. Otherwise $(1\,2\,\dotsb\,n)$ isn't even an element of $A_n$. To answer your question for odd $n$, you can reduce this to the fact you already know: that the collection of all $3$-cycles generate $A_n$.

1. Note that $(1\,2\,\dotsb\,n)(1\,2\,3)(1\,2\,\dotsb\,n)^{-1} = (2\,3\,4)$. Iteratively conjugating by $(1\,2\,\dotsb\,n)$ you can get all $3$-cycles of the form $(a\!-\!1\;\;\;a\;\;\;a\!+\!1)$. Let's call that a cyclic cycle.
2. Then you can do the sort of calculations that you were doing to write an arbitrary $3$-cycle as a product of these cyclic $3$-cycles. ... although actually writing this down rigorously turns out to be kinda messy.