Alternating Group is Simple except on 4 Letters/Lemma 3

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

Let $i, j, k \in$ $\N_n$ be such that $\rho = (i, j, k)$.

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


Proof

We will proceed by cases.

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


Case $1$
$\card {\{1,2,3\} \cap \{i, j, k\}} = 0$ (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 $2$
$\card {\{1,2,3\} \cap \{i, j, k\}} = 1$.

Without loss of generality, suppose that $ \{1,2,3\} \cap \{i, j, k\} = \{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, without loss of generality, 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 $3$
$\card {\{1,2,3\} \cap \{i, j, k\}} = 2$.

Without loss of generality, assume that $\{1,2,3\} \cap \{i, j, k\} = \{1, 2\}$.

We have three further sub-cases: either $i = 1 \land j = 2$, or $(i = 1 \land j \ne 2) \lor (i \ne 1 \land j = 2)$, 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 $\{1, 2, 3\}$.)

Suppose $i = 1 \land j = 2$.

Since $n \ge 5$, there exists an element $l \in \N_n \setminus \{1, 2, 3, k\}$.

Then, the permutation $\sigma = \tuple {3, k, l}$ is even (by Parity of K-Cycle) and has the desired property.

Now suppose $(i = 1 \land j \ne 2) \lor (i \ne 1 \land j = 2)$.

Without loss of generality, assume $(i = 1 \land j \ne 2)$.

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 $i = 2$ and $j = 1$, 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 $i = 2$ and $k = 1$.

Since $n \ge 5$, there exists an element $l \in \N_n \setminus \{1, 2, 3, k\}$.

Then, the permutation $\sigma = \tuple {1, 2, j, l, 3}$ is even and has the desired property.


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

That is, $\{i, j, k\} = \{1, 2, 3\}$.

We have that any (not necessarily even) permutation $\sigma$ has the desired property if and only if 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.

$\blacksquare$