A Group Action of the Symmetric Group SX on the Set X

# A Group Action of the Symmetric Group SX on the Set X

Let $X$ be a (nonempty) set and let $(S_X, \circ)$ be the symmetric group of $X$, i.e., the group of all permutations of the set $X$ with the operation of composition.

Define a left group action of the group $S_X$ on $X$ for all $\sigma \in S_X$ and for all $x \in X$ by:

(1)
\begin{align} \quad \sigma x = \sigma(x) \end{align}

Indeed, the map $(\sigma, x) \to \sigma (x)$ is a map from $S_X \times X \to X$. We will now verify that it is a left group action of $S_X$ on the set $X$.

For all $\sigma_1, \sigma_2 \in S_X$ and for all $x \in X$ we have that:

(2)
\begin{align} \quad \sigma_1 (\sigma_2x) = \sigma_1(\sigma_2(x)) = (\sigma_1 \circ \sigma_2)(x) = (\sigma_1 \circ \sigma_2)x \end{align}

And for all $x \in X$ we have that:

(3)
\begin{align} \quad \mathrm{id}_X x = \mathrm{id}_X(x) = x \end{align}

So indeed $(\sigma, x) \to \sigma(x)$ is a left group action of $S_X$ on the set $X$.

Let $\varphi : S_X \to S_X$ be the associated permutation representation homomorphism defined for all $\delta \in S_X$ by $\varphi(\delta) = \sigma_{\delta}$ where $\sigma_{\delta} : X \to X$ is defined for all $x \in X$ by $\sigma_{\delta}(x) = \delta x = \delta(x)$.

Observe that then $\varphi(\delta) = \delta$ for all $\delta \in S_X$, and so $\varphi : S_X \to S_X$ is simply the identity homomorphism of $S_X$ to $S_X$.