Alternating Group is Simple except on 4 Letters/Lemma 2

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 the permutation on $\N_n$ in the form:

$\alpha = \tuple {1, 2} \tuple {3, 4}$

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

Then:

$\beta^{-1} \alpha^{-1} \beta \alpha = \beta$

Proof

\(\ds \beta^{-1} \alpha^{-1} \beta \alpha\)

\(=\)

\(\ds \paren {3, 5, 4} \tuple {1, 2} \tuple {3, 4} \tuple {3, 4, 5} \tuple {1, 2} \tuple {3, 4}\)

\(\ds \)

\(=\)

\(\ds \paren {3, 5, 4} \tuple {1, 2} \tuple {3, 4} \tuple {1, 2} \tuple {3, 5}\)

\(\ds \)

\(=\)

\(\ds \paren {3, 5, 4} \tuple {3, 5, 4}\)

\(\ds \)

\(=\)

\(\ds \paren {3, 4, 5}\)

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 83$