Symmetric group

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Note: the symmetric group is a permutation group on finitely many symbols, see permutation group (which uses the same notation) for the more general case.

Definition

Let [ilmath]k\in\mathbb{N} [/ilmath] be given. The symmetric group on [ilmath]k[/ilmath] symbols, denoted [ilmath]S_k[/ilmath], is the permutation group on [ilmath]\{1,2,\ldots,k-1,k\}\subset\mathbb{N} [/ilmath]. The set of the group is the set of all permutations on [ilmath]\{1,2,\ldots,k-1,k\} [/ilmath]. See proof that the symmetric group is actually a group for details.

• Identity element: [ilmath]e:\{1,\ldots,k\}\rightarrow\{1,\ldots,k\} [/ilmath] which acts as so: [ilmath]e:i\mapsto i[/ilmath] - this is the identity permutation, it does nothing.
• The group operation is ordinary function composition, for [ilmath]\sigma,\tau\in S_k[/ilmath] we define:
• [ilmath]\sigma\tau:\eq \sigma\circ\tau[/ilmath] with: [ilmath]\sigma\tau:\{1,\ldots,k\}\rightarrow\{1,\ldots,k\} [/ilmath] by [ilmath]\sigma\tau:i\mapsto\sigma(\tau(i))[/ilmath]
• Caveat:Be careful, a lot of authors (Allenby, McCoy to name 2) go left-to-right and write [ilmath]i\sigma[/ilmath] for what we'd use [ilmath]\sigma(i)[/ilmath] or [ilmath]\sigma i[/ilmath] at a push for. Then [ilmath]\sigma\tau[/ilmath] would be [ilmath]\tau\circ\sigma[/ilmath] in our notation

Permutation notation

The "base" way to write a permutation, [ilmath]\sigma\in S_k[/ilmath], is as a table:

• $\left(\begin{array}{ccccc}1 & 2 & \cdots & k-1 & k\\ \sigma(1) & \sigma(2) & \cdots & \sigma(k-1) & \sigma(k)\end{array}\right)$

The top row is an element of the domain of [ilmath]\sigma[/ilmath] considered as a function and thing below it is the image of that element under [ilmath]\sigma[/ilmath]

This notation quickly becomes heavy so we switch to cycle notation, which we demonstrate below.

Note, however, that in order to use cycle notation we require the following:

Example

Let us consider [ilmath]S_5[/ilmath] as an example.

• Let [ilmath]\sigma\in S_5[/ilmath] be the permutation given as follows:
• [ilmath]\sigma:1\mapsto 3[/ilmath], [ilmath]\sigma:2\mapsto 2[/ilmath], [ilmath]\sigma:3\mapsto 5[/ilmath], [ilmath]\sigma:4\mapsto 1[/ilmath], [ilmath]\sigma:5\mapsto 4[/ilmath]
This can be written more neatly as:
• $\left(\begin{array}{ccccc}1 & 2 & 3 & 4 & 5 \\ 3 & 2 & 5 & 1 & 4\end{array}\right)$, the thing in the top row is sent to the thing below it.
• This can be written as the product of disjoint cycles too:
• $(1\ 3\ 5\ 4)$ or [ilmath](1\ 3\ 5\ 4)(2)[/ilmath] if you do not take the "implicit identity" part. That is any element not in a cycle stays the same
• Or as transpositions
• $(1\ 4)(1\ 5)(1\ 3)$ - recall we read right-to-left, so this is read:
• [ilmath]1\mapsto 3\mapsto 3\mapsto 3[/ilmath]
• [ilmath]3\mapsto 1\mapsto 5\mapsto 5[/ilmath]
• [ilmath]5\mapsto 5\mapsto 1\mapsto 4[/ilmath]
• [ilmath]4\mapsto 4\mapsto 4\mapsto 1[/ilmath] - the cycle [ilmath](1\ 3\ 5\ 4)[/ilmath]
• And of course [ilmath]2\mapsto 2\mapsto 2\mapsto 2[/ilmath]