Exercise 1.7.9 ($S_A$ acts on $A^k$)

Proof. If $X=(a_1,a_2,\dots ,a_k)\in A^k$, then $\sigma (X)=(\sigma (a_1),\sigma (a_2),\dots ,\sigma (a_k))\in A^k$.

(a)

If $\sigma ,\tau$ are permutations in $S_A$, and $X=(a_1,a_2,\dots ,a_k)\in A^k$, then

• $$\begin{array}{llll}\hfill \sigma \cdot (\tau \cdot X)& =\sigma \cdot (\tau \cdot (a_1,a_2,\dots ,a_k))& \hfill & \\ \hfill & =\sigma \cdot (\tau (a_1),\tau (a_2),\dots ,\tau (a_k))& \hfill & \\ \hfill & =(\sigma (\tau (a_1)),\sigma (\tau (a_2)),\dots ,\sigma (\tau (a_k)))& \hfill & \\ \hfill & =(\sigma \circ \tau )\cdot (a_1,a_2,\dots ,a_k)& \hfill & \\ \hfill & =(\sigma \circ \tau )\cdot X,& \hfill & \end{array}$$

• If $e={\mathrm{Id}}_A$, then

  $$\begin{array}{llll}\hfill e\cdot X& ={\mathrm{Id}}_A\cdot (a_1,a_2,\dots ,a_k)& \hfill & \\ \hfill & =({\mathrm{Id}}_A(a_1),{\mathrm{Id}}_A(a_2),\dots ,{\mathrm{Id}}_A(a_k))& \hfill & \\ \hfill & =(a_1,a_2,\dots ,a_k)& \hfill & \\ \hfill & =X.& \hfill & \end{array}$$

Therefore $S_A$ acts on the set $B$ by $\sigma \cdot \{a_1,\dots ,a_k\}=\{\sigma (a_1),\dots ,\sigma (a_k)\}$.

(b)

For instance, $S_3$ acts on ${[[1,4]]}^2$: $$\begin{array}{llll}\hfill (12)\cdot (1,1)& =(2,2),& \hfill & \\ \hfill (12)\cdot (1,2)& =(2,1),& \hfill & \\ \hfill (12)\cdot (1,3)& =(2,3),& \hfill & \\ \hfill (12)\cdot (1,4)& =(2,4),& \hfill & \\ \hfill (12)\cdot (2,1)& =(1,2),& \hfill & \\ \hfill (12)\cdot (2,2)& =(1,1),& \hfill & \\ \hfill (12)\cdot (2,3)& =(1,3),& \hfill & \\ \hfill (12)\cdot (2,4)& =(1,4),& \hfill & \\ \hfill (12)\cdot (3,1)& =(3,2),& \hfill & \\ \hfill (12)\cdot (3,2)& =(3,1),& \hfill & \\ \hfill (12)\cdot (3,3)& =(3,3),& \hfill & \\ \hfill (12)\cdot (3,4)& =(3,4),& \hfill & \\ \hfill (12)\cdot (4,1)& =(4,2),& \hfill & \\ \hfill (12)\cdot (4,2)& =(4,1),& \hfill & \\ \hfill (12)\cdot (4,3)& =(4,3),& \hfill & \\ \hfill (12)\cdot (4,4)& =(4,4),& \hfill & \\ \hfill & & \hfill \end{array}$$

As expected, $(12)$ induces a permutation on ${[[1,4]]}^2$.

Similarly:

$$\begin{array}{llll}\hfill (123)\cdot (1,1)& =(2,2)& \hfill & \\ \hfill (123)\cdot (1,2)& =(2,3)& \hfill & \\ \hfill (123)\cdot (1,3)& =(2,1),& \hfill & \\ \hfill \text{…}& & \hfill \end{array}$$