39 lines
2.2 KiB
Markdown
39 lines
2.2 KiB
Markdown
|
# Permutations
|
||
|
|
|
||
|
|
## Definition
|
||
|
|
|
||
|
|
> *Definition*: let $X$ be a set.
|
||
|
|
>
|
||
|
|
> * A bijection of $X$ to itself is called a permutation of $X$. The set of all permutations of $X$ is denoted by $\text{Sym}(X)$ and is called the symmetric group on $X$.
|
||
|
|
> * The product $g \cdot h$ of two permutations $g,h$ in $\text{Sym}(X)$ is defined as the composition $g \circ h$ of $g$ and $h$.
|
||
|
|
> * If $X = \{1, \dots, n\}$ we write $\mathrm{Sym}_n(X)$ instead of $\mathrm{Sym}(X)$.
|
||
|
|
|
||
|
|
<br>
|
||
|
|
|
||
|
|
> *Definition*: the identity map is defined as $\mathrm{id}: X \to X$ with $g = g \cdot \mathrm{id} = \mathrm{id} \cdot g$ for all $g$ in $\mathrm{Sym}(X)$. The inverse of $g$ denoted by $g^{-1}$ satisfies $g^{-1} \cdot g = g \cdot g^{-1} = \mathrm{id}$.
|
||
|
|
|
||
|
|
In matrix notation: let $g = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1\end{pmatrix}$ and $h = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3\end{pmatrix}$ with $g,h \in \mathrm{Sym}_3(X)$, then we can take
|
||
|
|
|
||
|
|
$$
|
||
|
|
g \cdot h = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ \hline 2 & 1 & 3 \\ 3 & 2 & 1\end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1\end{pmatrix},
|
||
|
|
$$
|
||
|
|
|
||
|
|
and we have $g^{-1} = \begin{pmatrix} 2 & 3 & 1 \\1 & 2 & 3 \end{pmatrix}$.
|
||
|
|
|
||
|
|
<br>
|
||
|
|
|
||
|
|
> *Theorem*: $\mathrm{Sym}_n$ has exactly $n!$ elements.
|
||
|
|
|
||
|
|
??? note "*Proof*:"
|
||
|
|
|
||
|
|
A permutation can be described in a matrix notation by a $2$ by $n$ matrix with the numbers $1,\dots,n$ in the first row and the images in the second row. There are $n!$ possibilities to fill the second row.
|
||
|
|
|
||
|
|
We can also omit the matrix notation and use the list notation for permutations then we have for $g = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1\end{pmatrix} = [2,3,1]$, as the first row speaks for itself.
|
||
|
|
|
||
|
|
<br>
|
||
|
|
|
||
|
|
> *Definition*: the order of a permutation $g$ is the smallest positive integer $m$ such that $g^m = \mathrm{id}$.
|
||
|
|
|
||
|
|
For example the order of the permutation $[2,1,3]$ in $\mathrm{Sym}_3$ is 2.
|
||
|
|
|
||
|
|
If $g$ is a permutation in $\mathrm{Sym}_n$ then the permutations $g, g^2, g^3, \dots$ can not all be distinct, since there are only $n!$ distinct permutations in $\mathrm{Sym}_n$. So there must exists a $r < s$ such that $g^r = g^s$. Since $g$ is a bijection there must be $g^{s-r} = e$. So there exist positive numbers $m$ with $g^m = e$ and in particular a smallest such number. Therefore each permutation $g$ has a well-defined order.
|