Exercise 2.2.12 (Action of $S_4$ on $\mathbb{Z}[x_1,x_2,x_3,x_4]$ and some stabilizers)

Proof. Consider the ring $R=\mathbb{Z}[x_1,x_2,x_3,x_4]$, and the mapping

(a)

If $\sigma =(1234)$ and $\tau =(123)$, then $\sigma \circ \tau =(1342)$ (we compose from right to left). Then $$\begin{array}{llll}\hfill \sigma \cdot p& =12x_1{x}_2^5{x}_3^7-18x_3^4{x}_4+11x_1^{23}{x}_2^6{x}_3{x}_4^3,& \hfill & \\ \hfill \tau \cdot (\sigma \cdot p)& =12x_1^7{x}_2{x}_3^5-18x_1^4{x}_4+11x_1{x}_2^{23}{x}_3^6{x}_4^7,& \hfill & \\ \hfill (\sigma \circ \tau )\cdot p& =12x_3^5{x}_1^7{x}_2-18x_1^3{x}_4+11x_3^6{x}_1{x}_4^3{x}_2^{23}& \hfill & \\ \hfill & =12x_1^7{x}_2{x}_3^5-18x_1^4{x}_4+11x_1{x}_2^{23}{x}_3^6{x}_4^3.& \hfill & \end{array}$$

Therefore $\tau \cdot (\sigma \cdot p)=(\sigma \circ \tau )\cdot p$.

(b)

We generalize this example. First $1\cdot p=p$. For all $\sigma ,\tau$ in $S_4$, if $j=\tau (i)$ then $\sigma \cdot x_{\tau (i)}=\sigma \cdot x_j=x_{\sigma (j)}=x_{\sigma (\tau (i))}$, therefore $$\begin{array}{lll}\hfill \sigma \cdot (\tau \cdot x_i)=\sigma \cdot x_{\tau (i)}=x_{\sigma (\tau (i))}=x_{(\mathit{\sigma \tau })(i)}=(\mathit{\sigma \tau })\cdot (x_i)(i=1,2,3,4).& & \hfill \text{(1) }\end{array}$$

By definition

$$\sigma \cdot p(x_1,x_2,x_3,x_4)=p(x_{\sigma (1)},x_{\sigma (2)},x_{\sigma (3)},x_{\sigma (4)}),$$

then

$$\begin{array}{lll}\hfill \sigma \cdot p(x_1,x_2,x_3,x_4)=p(\sigma \cdot x_1,\sigma \cdot x_2,\sigma \cdot x_3,\sigma \cdot x_4).& & \hfill \text{(2)}\end{array}$$

Therefore

$$\begin{array}{llllll}\hfill \sigma \cdot (\tau \cdot p)& =\sigma \cdot (\tau \cdot p(x_1,x_2,x_3,x_4))& \hfill & & \hfill & \\ \hfill & =\sigma \cdot p(\tau \cdot x_1,\tau \cdot x_2,\tau \cdot x_3,\tau \cdot x_4)& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =p(\sigma \cdot (\tau \cdot x_1),\sigma \cdot (\tau \cdot x_2),\sigma \cdot (\tau \cdot x_3),\sigma \cdot (\tau \cdot x_4))& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =p((\mathit{\sigma \tau })\cdot x_1,(\mathit{\sigma \tau })\cdot x_2,(\mathit{\sigma \tau })\cdot x_3,(\mathit{\sigma \tau })\cdot x_4)& \hfill \text{(by (1))}& & \hfill & & \hfill \\ \hfill & =(\mathit{\sigma \tau })\cdot p(x_1,x_2,x_3,x_4)& \hfill \text{(by (2))}& & \hfill & & \hfill \\ \hfill & =(\mathit{\sigma \tau })\cdot p.& \hfill & & \hfill & \end{array}$$

So $\phi$ is a left action of $S_4$ on $R$

(c)

Let $G$ denote the group $S_4$.

Note that $\sigma \in S_4$ stabilizes $x_4$ if and only if $x_4=\sigma \cdot x_4=x_{\sigma (4)}$, if and only if $4=\sigma (4)$, i.e., $\sigma \in G_4$, where $G_4$ is the stabilizer of $4$ for the action of $S_4$ on $\{1,2,3,4\}$. So the stabilizer $G_{x_4}$ of $x_4$ is $G_{x_4}=G_4$

We proved in Exercise 8 that

$$\phi \{\begin{array}{ccc}\hfill G_4\hfill & \hfill \text{↦}\hfill & S_{[[1,3]]}=S_3\hfill \\ \hfill \sigma \hfill & \hfill \text{↦}\hfill & \sigma {\text{|}}_{[[1,3]]},\hfill \end{array}$$

is an isomorphism. So

$$G_{x_4}=G_4=\{(),(12),(13),(23),(123),(132)\}\subset S_4$$

is isomorphic to $S_3$.

(d)

Let $G_{x_1+x_2}$ be the stabilizer of $x_1+x_2$ for the action $\phi$. Then $$\begin{array}{llll}\hfill \sigma \in G_{x_1+x_2}& ⟺\sigma \cdot (x_1+x_2)=x_1+x_2& \hfill & \\ \hfill & ⟺x_{\sigma (1)}+x_{\sigma (2)}=x_1+x_2& \hfill & \\ \hfill & ⟺(\sigma (1)=1\text{ and }\sigma (2)=2)\text{ or }(\sigma (1)=2\text{ and }\sigma (2)=1)& \hfill & \\ \hfill & ⟺\sigma \in \left\{\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right),\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill & \hfill 3\hfill \end{array}\right),\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right),\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 4\hfill & \hfill 3\hfill \end{array}\right)\right\}& \hfill & \\ \hfill & ⟺\sigma \in \{(),(34),(12),(12)(34)\},& \hfill & \end{array}$$

so

$$G_{x_1+x_2}=\{(),(34),(12),(12)(34)\}.$$

Since a stabilizer is a subgroup, $V=\{(),(34),(12),(12)(34)\}$ is a subgroup of $S_4$ of order $4$.

(Since all the elements of $V$ are of order $1$ or $2$, $V\simeq V_4=Z_2\times Z_2$.)

(e)

Similarly, $$\begin{array}{llll}\hfill \sigma \in G_{x_1{x}_2+x_3{x}_4}⟺& \sigma \cdot (x_1{x}_2+x_3{x}_4)=x_1{x}_2+x_3{x}_4& \hfill & \\ \hfill ⟺& x_{\sigma (1)}{x}_{\sigma (2)}+x_{\sigma (3)}{x}_{\sigma (4)}=x_1{x}_2+x_3{x}_4& \hfill & \\ \hfill ⟺& (x_{\sigma (1)}{x}_{\sigma (2)}=x_1{x}_2\text{ and }{x}_{\sigma (3)}{x}_{\sigma (4)}=x_3{x}_4)& \hfill & \\ \hfill & \text{ or }& \hfill & \\ \hfill & (x_{\sigma (1)}{x}_{\sigma (2)}=x_3{x}_4\text{ and }{x}_{\sigma (3)}{x}_{\sigma (4)}=x_1{x}_2)& \hfill & \\ \hfill ⟺& \sigma \in \left\{\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right),\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill & \hfill 3\hfill \end{array}\right),\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right),\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 4\hfill & \hfill 3\hfill \end{array}\right)\right\}& \hfill & \\ \hfill & \cup & \hfill & \\ \hfill & \left\{\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right),\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right),\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right),\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right)\right\}& \hfill & \\ \hfill & & \hfill \end{array}$$

so

$$\begin{array}{lll}\hfill G_{x_1{x}_2+x_3{x}_4}=\{(),(34),(12),(12)(34),(13)(24),(1324),(1423),(14)(23)\}(=W).& & \hfill \end{array}$$

( $W$ is a subgroup of $S_4$ because it is a stabilizer : we can view $W$ as the group of symmetries of a square whose vertices are numbered 1, 3, 2, 4)

More concisely,

$$G_{x_1{x}_2+x_3{x}_4}=W=⟨(1324),(14)(23)⟩.$$

If we put $\rho =(1324)$ and $\sigma =(14)(23)$, then $W=⟨\rho ,\sigma ⟩$ and ${\rho }^4={\sigma }^2=e$ and $\mathit{\rho \sigma }=\sigma {\rho }^{-1}$, so there is a surjective homomorphism

$$\phi :D_8=⟨r,s\mid r^4=s^2=e,\mathit{rs}=s{r}^{-1}⟩\to W$$

which maps $r$ on $\rho$ and $s$ on $\sigma$. Moreover $|D_8|=|W|=8$. Therefore $D_8\simeq W$.

In conclusion, $G_{x_1{x}_2+x_3{x}_4}\simeq D_8$.

(f)

Let $\sigma$ be a permutation in $S_4=G$. $$\begin{array}{llllll}\hfill \sigma \in G_{(x_1+x_2)(x_3+x_4)}& ⟺\sigma \cdot (x_1+x_2)(x_3+x_4)=(x_1+x_2)(x_3+x_4)& \hfill & & \hfill & \\ \hfill & ⟺(x_{\sigma (1)}+x_{\sigma (2)})(x_{\sigma (3)}+x_{\sigma (4)})=(x_1+x_2)(x_3+x_4)& \hfill & & \hfill & \\ \hfill ⟺& (x_{\sigma (1)}+x_{\sigma (2)}=x_1+x_2\text{ and }{x}_{\sigma (3)}+x_{\sigma (4)}=x_3{x}_4)& \hfill & & \hfill & \\ \hfill & \text{ or }& \hfill & & \hfill & \\ \hfill & (x_{\sigma (1)}+x_{\sigma (2)}=x_3+x_4\text{ and }{x}_{\sigma (3)}+x_{\sigma (4)}=x_1+x_2)& \hfill & & \hfill & \\ \hfill ⟺& (x_{\sigma (1)}{x}_{\sigma (2)}=x_1{x}_2\text{ and }{x}_{\sigma (3)}{x}_{\sigma (4)}=x_3{x}_4)& \hfill & & \hfill & \\ \hfill & \text{ or }& \hfill & & \hfill & \\ \hfill & (x_{\sigma (1)}{x}_{\sigma (2)}=x_3{x}_4\text{ and }{x}_{\sigma (3)}{x}_{\sigma (4)}=x_1{x}_2)& \hfill & & \hfill & \\ \hfill ⟺& \sigma \in G_{x_1{x}_2+x_3{x}_4}& \hfill \text{(by part (e))}.& & \hfill & & \hfill \end{array}$$

So

$$G_{(x_1+x_2)(x_3+x_4)}=G_{x_1{x}_2+x_3{x}_4}.$$