# Group Action of Symmetric Group on Subset is Transitive

## 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:

$\forall \pi \in S_n, \forall S \in B_r: \pi * B_r = \pi \sqbrk S$

where $\pi \sqbrk S$ denotes the image of $S$ under $\pi$.

Then $*$ is a transitive group action.

## Proof

From Group Action of Symmetric Group on Subset it is established that $*$ is a group action.

Let $U = \set {u_1, u_2, \ldots, u_r}$ and $V = \set {v_1, v_2, \ldots, v_r}$ be elements of $B_r$.

Then there exists a permutation $\rho \in S_n$ such that:

$\map \rho {u_k} = v_k$

for all $k \in \set {1, 2, \ldots, r}$.

Thus:

$\rho \sqbrk U = V$

Thus $B_r$ is the orbit of all $U \in B_r$.

Hence the result by definition of transitive group action.

$\blacksquare$