Group Action of Symmetric Group/Subset
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Theorem
Let $\N_n$ denote the set $\set {1, 2, \ldots, n}$.
Let $\struct {S_n, \circ}$ denote the symmetric group on $\N_n$.
Let $r \in \N: 0 < r \le n$.
Let $B_r$ denote the set of all subsets of $\N_n$ of cardinality $r$:
- $B_r := \set {S \subseteq \N_n: \card S = r}$
Let $*$ be the mapping $*: S_n \times B_r \to B_r$ defined as:
- $\forall \pi \in S_n, \forall S \in B_r: \pi * S = \pi \sqbrk S$
where $\pi \sqbrk S$ denotes the image of $S$ under $\pi$.
Then $*$ is a group action.
Proof
The group action axioms are investigated in turn.
Let $\pi, \rho \in S_n$.
Let $S \in B_r$.
Thus:
\(\ds \pi * \paren {\rho * S}\) | \(=\) | \(\ds \pi * \rho \sqbrk S\) | Definition of $*$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \pi \sqbrk {\rho \sqbrk S}\) | Definition of $*$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\pi \circ \rho} \sqbrk S\) | Definition of Composition of Mappings | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\pi \circ \rho} * n\) | Definition of $*$ |
demonstrating that Group Action Axiom $\text {GA} 1$ holds.
Then:
\(\ds I_{\N_n} * S\) | \(=\) | \(\ds I_{\N_n} \sqbrk S\) | where $I_{\N_n}$ is the identity mapping on $\N_n$ | |||||||||||
\(\ds \) | \(=\) | \(\ds S\) | Definition of Identity Mapping |
demonstrating that Group Action Axiom $\text {GA} 2$ holds.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 6.5$. Orbits: Example $120$