Show that the action of the alternating group $A_n$ on $\{1,2,\dots,n\}$ is $(n-2)$ - transitive.
Proof: Suppose that $I,J\subset \{1,2,\dots,n\}$ with $|I|=|J|=n-2$ and $I=\{x_1,x_2,\dots,x_{n-2}\}$ and $J=\{y_1,y_2,\dots,y_{n-2}\}$.
Case 1. Suppose that $I=J$ but possibly elemens of $I$ and $J$ are arranged in some order. Consider sets $$A=\{1\leq i\leq n-2: x_i=y_i\} \ \text{and} \ B=\{1\leq i\leq n-2: x_i\neq y_i\}$$ Then consider $\sigma:I\to J$ defined by $\sigma=\prod\limits_{i\in B}(x_i,y_i)$ and it's easy to see that $\sigma(x_i)=y_i$ for all $1\leq i \leq n-2$. If $|B|$ - even then $\sigma\in A_n$ but if $|B|$ - odd then replace $\sigma$ by $\sigma (bc)$ where $b,c\in \{1,2,\dots,n\}-\{x_1,\dots,x_n\}$.
Case 2. Suppose that $I$ and $J$ differs by one element. It means $\exists a\in I-J$ and $\exists b\in J-I$.
2.1. Suppose that $x_k=a$ and $y_k=b$ then consider sets $$A=\{1\leq i\leq n-2,i\neq k: x_i=y_i\} \ \text{and} \ B=\{1\leq i\leq n-2,i\neq k: x_i\neq y_i\}$$ Then consider $\sigma=\prod \limits_{i\in B}(x_i,y_i)(a,b)$ and if $|B|$-odd then $\sigma\in A_n$ and if $|B|$ is even then replace $\sigma$ by $\sigma(c,b)$ where $c\in \{1,\dots,n\}$ different from elements of $I$ and $J$.
2.2. If $x_k=a$ and $y_k\neq b$. Then take $\sigma=\prod \limits_{i\in B}(x_i,y_i)(a,y_k)$. If $|B|$-odd then $\sigma \in A_n$ and if $|B|-even$ then replace $\sigma$ by $\sigma(b,c)$.
However I am not able to apply this argument to the case when $I$ and $J$ differs by two elements.
Remark: Please do not duplicate this topic since in existing topics i was not able to understand the solution completely.
EDIT: I solved the case when sets differs only by one element.
Can anyone show how to solve the third cases? Is my reasoning above correct?
