# 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:

 $\ds \map {\beta^{-1} \alpha^{-1} \beta \alpha} 1$ $=$ $\ds \map {\paren {3, 5, 4} \alpha^{-1} \tuple {3, 4, 5} } {\map \alpha 1}$ $\ds$ $=$ $\ds \map {\paren {3, 5, 4} \alpha^{-1} \tuple {3, 4, 5} } 2$ $\ds$ $=$ $\ds \map {\paren {3, 5, 4} \alpha^{-1} } 2$ as $2$ is fixed by $\tuple {3, 4, 5}$ $\ds$ $=$ $\ds \map {\paren {3, 5, 4} } 1$ as $\map \alpha 1 = 2$ $\ds$ $=$ $\ds 1$ as $1$ is fixed by $\tuple {3, 4, 5}$

$\blacksquare$