Alternating Group is Simple except on 4 Letters/Lemma 3

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

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

Let $\rho \in S_n$ be an arbitrary $3$-cycle.

Let $\N_n$ denote the initial segment of the natural numbers:
 * $\set {0, 1, \ldots, n - 1}$

Let $i, j, k \in \N_n$ be such that $\rho = \tuple {i, j, k}$

Then there exists an even permutation $\sigma \in A_n$ such that $\map \sigma 1 = i$, $\map \sigma 2 = j$ and $\map \sigma 3 = k$

Proof
We will proceed by cases.

We have that $\card {\set {1, 2, 3} \cap \set {i, j, k} }$ is either $0$, $1$, $2$ or $3$


 * Case $1$: $\card {\set {1, 2, 3} \cap \set {i, j, k} } = 0$ (this case is only possible when $n \ge 6$)

The permutation：
 * $\sigma = \tuple {1, i, 2, j} \tuple {3, k}$ is even (by Parity of K-Cycle and Sign of Composition of Permutations) and has the desired property.


 * Case $2$: $\card {\set {1, 2, 3} \cap \set {i, j, k} } = 1$

, suppose that：
 * $\set {1, 2, 3} \cap \set {i, j, k} = \set 1$

We have two further sub-cases:
 * $i = 1$

or：
 * $i \ne 1$

If
 * $i = 1$

then, the permutation：
 * $\sigma = \tuple {2, j} \tuple {3, k}$

is even (by Parity of K-Cycle and Sign of Composition of Permutations) and has the desired property.

Now suppose:
 * $i \ne 1$

, assume that:
 * $j = 1$

Then the permutation:
 * $\sigma = \tuple {1, i, 3, k, 2}$

is even (by Parity of K-Cycle) and has the desired property.


 * Case $3$: $\card {\set {1, 2, 3} \cap \set {i, j, k} } = 2$

, assume that:
 * $\set {1, 2, 3} \cap \set {i, j, k} = \set {1, 2}$

We have three further sub-cases:
 * $i = 1 \land j = 2$
 * $\paren {i = 1 \land j \ne 2} \lor \paren {i \ne 1 \land j = 2}$

or:
 * $i \ne 1 \land j \ne 2$

(In other words, if $\sigma$ exists, it either fixes $2$, $1$, or $0$ letters of those in $\set {1, 2, 3}$)

Suppose:
 * $i = 1 \land j = 2$

Since:
 * $n \ge 5$

there exists an element:
 * $l \in \N_n \setminus \set {1, 2, 3, k}$

Then, the permutation:
 * $\sigma = \tuple {3, k, l}$

is even (by Parity of K-Cycle) and has the desired property.

Now suppose:
 * $\paren {i = 1 \land j \ne 2} \lor \paren {i \ne 1 \land j = 2}$

, assume:
 * $\paren {i = 1 \land j \ne 2}$

This implies:
 * $k = 2$

Then, the permutation:
 * $\sigma = \tuple {2, j, 3}$

is even (by Parity of K-Cycle) and has the desired property.

Now suppose:
 * $i \ne 1 \land j \ne 2$

If:
 * $i = 2$ \land $j = 1$

then the permutation:
 * $\sigma = \tuple {1, 2} \tuple {3, k}$

is even (by Parity of K-Cycle and Sign of Composition of Permutations) and has the desired property.

Otherwise, without loss of generality, assume:
 * $i = 2$ \land $k = 1$

Since:
 * $n \ge 5$

there exists an element:
 * $l \in \N_n \setminus \set {1, 2, 3, k}$

Then, the permutation:
 * $\sigma = \tuple {1, 2, j, l, 3}$

is even and has the desired property.


 * Case $4$: $\card {\set {1, 2, 3} \cap \set {i, j, k} } = 3$

That is:
 * $\set {i, j, k} = \set {1, 2, 3}$

We have that a (not necessarily even) permutation:
 * $\sigma$

has the desired property it can be written as a disjoint product:
 * $\sigma = \beta\alpha$

where $\alpha$ is the permutation:
 * $ \begin{pmatrix} 1 & 2 & 3 & 4 & \cdots & n \\ i & j & k & 4 & \cdots & n \\ \end{pmatrix}$

If $\alpha$ fixes all letters, $\alpha$ is the identity.

Take $\beta$ to be the identity.

Then $\sigma$ will be the identity permutation, so by Identity Mapping on Symmetric Group is Even Permutation, it will be even.

If $\alpha$ fixes one letter, $\alpha$ will be a transposition.

Since:
 * $n \ge 5$

we can take $\beta$ to be another transposition disjoint from $\alpha$

Then $\sigma$ will be the product of two disjoint transpositions.

Therefore, by Parity of K-Cycle and Sign of Composition of Permutations $\sigma$ will be even.

If $\alpha$ fixes no letters, $\alpha$ will be a $3$-cycle.

Take $\beta$ to be the identity permutation.

Then:
 * $\sigma = \alpha$

By Parity of K-Cycle, $\sigma$ will be even.

In all cases, we found an even permutation:
 * $\sigma$

with the desired property.

Also, $\rho$ was arbitrary.

Hence a $\sigma$ as described above always exists for any $3$-cycle.