Generators of alternating group $A_n$ for odd and even $n$

Question

Assume $n\geq3$ and let $A_n$ be the alternating group of $\{1,\ldots,n\}$. I would like to demonstrate the following claims:

1. $A_n=\langle(123),(12...n)\rangle$, if $n$ is odd;
2. $A_n=\langle(123),(23...n)\rangle$, if $n$ is even.

How can I use $A_n=\langle(123),(124),\ldots(12n)\rangle$ to prove these two claims? Can someone please help me get this started?

Let $n$ be odd. My idea was to show that $$\{(123),(124),\ldots(12n)\}\subset\langle(123),(12...n)\rangle.$$ I am not sure how to show that $$(12k)\in\langle(123),(12...n)\rangle$$ for $4\leq k\leq n$.

Answer 1

By repeatedly conjugating $(123)$ by $(123\ldots n)$, you obtain the transpositions $(123)$, $(234)$, $(345),\ldots,(n-2,n-1,n)$, $(n-1, n,1)$, and $(n,1,2)$ (and their inverses); in case you are unfamiliar, conjugation $x$ by $y$ for me means computing $yxy^{-1}$.

Once you have these, from $(123)$ you can get $(124)$ by taking, e.g., $(345)(123)(345)^{-1}$. Then $(125)$ by conjugating again; etc. You can easily obtain all $3$-cycles of the form $(12a)$, $3\leq a\leq n$.

Similarly for 2.