Even Permutation is Product of 3-Cycles
Theorem
Let $n \in \N$ such that $n \ge 3$.
Let $\pi: \N_n \to \N_n$ be a permutation on $n$ letters.
Let $\pi$ be an even permutation.
Then $\pi$ can be expressed as the composition of cyclic permutations of length $3$.
Proof
Let $A_n$ denote the alternating group on $n$ letters.
By Alternating Group is Set of Even Permutations, $A_n$ is the set of all even permutations of $S_n$.
$\pi$ can therefore be analysed in the context of elements of $A_n$.
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\pi \in A_n$ can be expressed as the composition of cyclic permutations of length $3$
From Identity Mapping on Symmetric Group is Even Permutation, the identity permutation $e$ on $S_n$ is even.
But there are no cyclic permutations of length $3$ in $S_1$ or $S_2$.
So $\map P 1$ and $\map P 2$ are false.
Basis for the Induction
$\map P 3$ is the case:
- $\pi \in A_3$ can be expressed as the composition of cyclic permutations of length $3$.
The even permutations of $A_3$ are $e$, $\tuple {1, 2, 3}$ and $\tuple {1, 3, 2}$.
Thus $\map P 3$ is seen to hold trivially.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that if $\map P {k - 1}$ is true, where $k > 3$, then it logically follows that $\map P k$ is true.
So this is the induction hypothesis:
- $\pi \in A_{k - 1}$ can be expressed as the composition of cyclic permutations of length $3$.
from which it is to be shown that:
- $\pi \in A_k$ can be expressed as the composition of cyclic permutations of length $3$.
Induction Step
This is the induction step:
Let $\pi \in A_n$.
Consider the permutation:
- $\sigma := \tuple {\pi_n, n, i} \circ \pi$
where $\pi_i = n$.
We have:
\(\ds \map \sigma n\) | \(=\) | \(\ds \tuple {\pi_n, n, i} \circ \map \pi n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\tuple {\pi_n, n, i} } {\pi_n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n\) |
and is even.
Because $\sigma$ fixes $n$ and is even, $\sigma$ is an even permutation of $\set {1, 2, \ldots, n - 1}$.
By the |inductive hypothesis, $\sigma$ is the composition of $3$-cycles, say:
- $\sigma = \alpha_1 \alpha_2 \dotsm \alpha s$
Setting $\alpha_0 := \tuple {\pi_n, i, n}$, we have:
\(\ds \alpha_0 \alpha_1 \alpha_2 \dotsm \alpha_s\) | \(=\) | \(\ds \alpha_0 \sigma\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {\pi_n, i, n} \tuple {\pi_n, n, i} \circ \pi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \pi\) |
and we have expressed $\pi$ as the composition of $3$-cycles
So $\map P {k - 1} \implies \map P k$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{\ge 0}: \pi \in A_n$ can be expressed as the composition of cyclic permutations of length $3$.
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 82$