Alternating Group is Simple except on 4 Letters/Lemma 1

Theorem
Let $n$ be an integer such that $n \ge 5$.

Let $A_n$ denote the alternating group on $n$ letters.

Let $\alpha \in A_n$ be a permutation on $\N_n$ such that $\map \alpha 1 = 2$.

Let $\beta$ be the $3$-cycle $\tuple {3, 4, 5}$.

Then the permutation $\beta^{-1} \alpha^{-1} \beta \alpha$ fixes $1$.

Proof
We let $\beta^{-1} \alpha^{-1} \beta \alpha$ act on $1$.

By construction:
 * $\map \alpha 1 = 2$

Thus: