Existence and Uniqueness of Cycle Decomposition

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Theorem

Let $S_n$ denote the symmetric group on $n$ letters.

Every element of $S_n$ may be uniquely expressed as a cycle decomposition, up to the order of factors.


Proof

By definition, a cycle decomposition of an element of $S_n$ is a product of disjoint cycles.


Construction of Disjoint Permutations

Let $\sigma \in S_n$ be a permutation on $S_n$.

Let $\mathcal R_\sigma$ be the equivalence defined in Permutation Induces Equivalence Relation.

Let $\N_n$ be used to denote the (one-based) initial segment of natural numbers:

$\N_n = \closedint 1 n = \set {1, 2, 3, \ldots, n}$

Let $\N_n / \mathcal R_\sigma = \set {E_1, E_2, \ldots, E_m}$ be the quotient set of $\N_n$ determined by $\mathcal R_\sigma$.


By Equivalence Class of Element is Subset:

$E \in \N_n / \mathcal R_\sigma \implies E \subseteq \N_n$

For any $E_i \in \N_n / \mathcal R_\sigma$, let $\rho_i: \paren {\N_n \setminus E_i} \to \paren {\N_n \setminus E_i}$ be the identity mapping on $\N_n \setminus E_i$.

By Identity Mapping is Permutation, $\rho_i$ is a permutation.


Also, let $\phi_i = \tuple {E_i, E_i, R}$ be a relation where $R$ is defined as:

$\forall x, y \in E_i: \tuple {x, y} \in R \iff \map \sigma x = y$


It is easily seen that $\phi_i$ is many to one.

For all $x \in E_i$:

\(\displaystyle x\) \(\mathcal R_\sigma\) \(\displaystyle \map \sigma x\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \sigma x\) \(\in\) \(\displaystyle E_i\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sigma \sqbrk {E_i}\) \(\subseteq\) \(\displaystyle E_i\) $\quad$ $\quad$

which shows that $\phi_i$ is left-total.


It then follows from the definition of a mapping that $\phi_i: E_i \to E_i$ is a mapping defined by:

$\map {\phi_i} x = \map \sigma x$

It is seen that $\phi_i$ is an injection because $\sigma$ is an injection.

So by Injection from Finite Set to Itself is Permutation, $\phi_i$ is a permutation on $E_i$.


By Intersection with Relative Complement is Empty, $E_i$ and $\N_n \setminus E_i$ are disjoint.

By Union with Relative Complement:

$E_i \cup \paren {\N_n \setminus E_i} = \N_n$

So by Union of Bijections with Disjoint Domains and Codomains is Bijection, let the permutation $\sigma_i \in S_n$ be defined by:

$\map {\sigma_i} x = \map {\paren {\phi_i \cup \rho_i} } x = \begin{cases} \map \sigma x & : x \in E_i \\ x & : x \notin E_i \end{cases}$


By Equivalence Classes are Disjoint, it follows that each of the $\sigma_i$ are disjoint.

$\Box$


These Permutations are Cycles

It is now to be shown that all of the $\sigma_i$ are cycles.


From Order of Element Divides Order of Finite Group, there exists $\alpha \in \Z_{\gt 0}$ such that $\sigma_i^\alpha = e$, and so:

$\map {\sigma_i^\alpha} x = \map e x = x$

By the Well-Ordering Principle, let $k = \min \set {\alpha \in \N_{\gt 0}: \map {\sigma_i^\alpha} x = x}$


Because $\sigma_i$ fixes each $y \notin E_i$, it suffices to show that:

$E_i = \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x}$

for some $x \in E_i$.


If $x \in E_i$, then for all $t \in \Z$:

$x \mathrel {\mathcal R_\sigma} \map {\sigma_i^t} x \implies \map {\sigma_i^t} x \in E_i$


It has been shown that:

$(1) \quad \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x} \subseteq E_i$


Let $x, y \in E_i$.

Then:

\(\displaystyle x\) \(\mathcal R_\sigma\) \(\displaystyle y\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {\sigma_i^t} x\) \(=\) \(\displaystyle y\) $\quad$ for some $t \in \Z$, by Permutation Induces Equivalence Relation $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {\sigma_i^{k q + r} } x\) \(=\) \(\displaystyle y\) $\quad$ for some $q \in \Z$, and $0 \le r \lt k$ by the Division Theorem $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {\sigma_i^r \sigma_i^{k q} } x\) \(=\) \(\displaystyle y\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {\sigma_i^r} x\) \(=\) \(\displaystyle y\) $\quad$ Fixed Point of Permutation is Fixed Point of Power $\quad$


It has been shown that:

$(2) \quad E_i \subseteq \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x}$


Combining $(1)$ and $(2)$ yields:

$E_i = \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x}$

$\Box$


The Product of These Cycles form the Permutation

Finally, it is now to be shown that $\sigma = \sigma_1 \sigma_2 \cdots \sigma_m$.


From the Fundamental Theorem on Equivalence Relations:

$x \in \N_n \implies x \in E_j$

for some $j \in \set {1, 2, \ldots, m}$.

Therefore:

\(\displaystyle \map {\sigma_1 \sigma_2 \cdots \sigma_m} x\) \(=\) \(\displaystyle \map {\sigma_1 \sigma_2 \cdots \sigma_j} x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map {\sigma_j} x\) $\quad$ because $\sigma_j \sqbrk {E_j} = E_j$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map \sigma x\) $\quad$ Definition of $\sigma_i$ $\quad$


and so existence of a cycle decomposition has been shown.

$\Box$


Uniqueness of Cycle Decomposition

Take the cycle decomposition of $\sigma$, which is $\sigma_1 \sigma_2 \cdots \sigma_m$.

Let $\tau_1 \tau_2 \cdots \tau_s$ be some product of disjoint cycles such that $\sigma = \tau_1 \tau_2 \cdots \tau_s$.

It is assume that this product describes $\sigma$ completely and does not contain any duplicate $1$-cycles.

Let $x$ be a moved element of $\sigma$.


Then there exists a $j \in \set {1, 2, \ldots, s}$ such that $\map {\tau_j} x \ne x$.

And so:

\(\displaystyle \map \sigma x\) \(=\) \(\displaystyle \map {\tau_1 \tau_2 \cdots \tau_j} x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map {\tau_j} x\) $\quad$ Power of Moved Element is Moved $\quad$


It has already been shown that $x \in E_i$ for some $i \in \set {1, 2, \ldots, m}$.

Therefore:

\(\displaystyle \map {\sigma_i} x\) \(=\) \(\displaystyle \map {\tau_j} x\) $\quad$ $\quad$
\(\displaystyle \map {\sigma_i^2} x\) \(=\) \(\displaystyle \map {\tau_{j \prime} \tau_j} x\) $\quad$ by Power of Moved Element is Moved $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map {\tau_j^2} x\) $\quad$ because $\map {\sigma_i^2} x \ne \map {\sigma_i} x$ and this product is disjoint $\quad$
\(\displaystyle \vdots\) \(\vdots\) \(\displaystyle \vdots\) $\quad$ $\quad$
\(\displaystyle \map {\sigma_i^{k - 1} } x\) \(=\) \(\displaystyle \map {\tau_j^{k - 1} } x\) $\quad$ $\quad$


This effectively shows that $\sigma_i = \tau_j$.


Doing this for every $E_i$ implies that $m = s$ and that there exists a $\rho \in S_m$ such that:

$\sigma_{\map \rho i} = \tau_i$


In other words, $\tau_1 \tau_2 \cdots \tau_m$ is just a reordering of $\sigma_1 \sigma_2 \cdots \sigma_m$.

$\blacksquare$


Also see


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