Alternating Group is Simple except on 4 Letters/Lemma 1

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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 $\tuple {3, 4, 5}$

$\blacksquare$


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