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 $\tuple {3, 4, 5}$ |
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 83$