## Exhibit an action of Sym(4) on a set of polynomials

Let $n$ be a positive integer and let $R$ be the set of all polynomials with integer coefficients in the independent variables $x_1, x_2, \ldots, x_n$. I.e., elements of $R$ are formal sums $\sum_{i \in I} a_i \prod_{j=1}^n x_i^{r_{i,j}}$ for some finite set $I$, integers $a_i$, and nonnegative integers $r_{i,j}$.

For each $\sigma \in S_n$, define a mapping $\sigma \cdot$ by $\sigma \cdot \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}}$ $= \sum_{i \in I} a_i \prod_{j=1}^n x_{\sigma(j)}^{r_{i,j}}$.

Prove that this defined a left group action of $S_4$ on $R$.

Let $p = \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}}$. We have $1 \cdot p = 1 \cdot \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}} = \sum_{i \in I} a_i \prod_{j=1}^n x_{\mathsf{id}(j)}^{r_{i,j}}$ $= \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}}$ $= p$.

Now let $\sigma, \tau \in S_n$. Then

 $\sigma \cdot (\tau \cdot p)$ = $\sigma \cdot (\tau \cdot \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}})$ = $\sigma \cdot \sum_{i \in I} a_i \prod_{j=1}^n x_{\tau(j)}^{r_{i,j}}$ = $\sum_{i \in I} a_i \prod_{j=1}^n x_{\sigma(\tau(j))}^{r_{i,j}}$ = $\sum_{i \in I} a_i \prod_{j=1}^n x_{(\sigma \circ \tau)(j)}^{r_{i,j}}$ = $(\sigma \circ \tau) \cdot \sum_{i \in I} a_i \prod_{j=1}^n x_j^{r_{i,j}}$

Thus we have a group action of $S_n$ on $R$.