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Sym(4) acts on a set of polynomials by permuting the variables


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.2 Exercise 2.2.12

Let $R$ be the set of all polynomials with integers coefficients in the independent variables $x_1, x_2, x_3, x_4$; i.e., elements of $R$ are formal sums $$p(x_1,x_2,x_3,x_4) = \sum_{i \in I} a_i \prod_{j=1}^4 x_j^{r_{i,j}}$$ for some finite set $I$, integers $a_i$, and nonnegative integers $r_{i,j}$.

Each $\sigma \in S_4$ permutes the$ x_i$ by $\sigma \cdot x_i = x_{\sigma(i)}$. We extend this action to a mapping $R \rightarrow R$ by $$\sigma \cdot p(x_1,x_2,x_3,x_4) = p(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)},x_{\sigma(4)}).$$ That is, $\sigma$ permutes the indices of the variables.

(1) Let $$p = p(x_1,x_2,x_3,x_4) = 12 x_1^5 x_2^7 x_4 - 18 x_2^3 x_3 + 11 x_1^6 x_2 x_3^3 x_4^{23},$$ and let $\sigma = (1\ 2\ 3\ 4)$ and $\tau = (1\ 2\ 3)$. Compute $\sigma \cdot p$, $\tau \cdot (\sigma \cdot p)$, $(\tau \circ \sigma) \cdot p$, and $(\sigma \circ \tau) \cdot p$.
(2) Prove that this set of mappings gives a group action of $S_4$ on $R$.
(3) Exhibit all permutations in $S_4$ that stabilize $x_4$ and prove that they form a subgroup isomorphic to $S_3$.
(4) Exhibit all permutations in $S_4$ that stabilize the element $x_1 + x_2$ and prove that they form an abelian group of order 4.
(5) Exhibit all permutations in $S_4$ that stabilize the element $x_1x_2 + x_3x_4$ and prove that they form a subgroup isomorphic to $D_8$.
(6) Show that the permutations in $S_4$ that stabilize the element $(x_1 + x_2)(x_3 + x_4)$ are exactly those found in the previous part.


Solution:(1)\begin{align*} &\ \sigma \cdot p\\
=&\ 12 x_2^5 x_3^7 x_1 - 18 x_3^3 x_4 + 11 x_2^6 x_3 x_4^3 x_1^{23}\\
=&\ 12 x_1 x_2^5 x_3^7 - 18 x_3^3 x_4 + 11 x_1^{23} x_2^6 x_3 x_4^3\end{align*}\begin{align*} &\ \tau \cdot (\sigma \cdot p)\\
=&\ \tau \cdot 12 x_1 x_2^5 x_3^7 - 18 x_3^3 x_4 + 11 x_1^{23} x_2^6 x_3 x_4^3\\
=&\ 12 x_2 x_3^5 x_1^7 - 18 x_1^3 x_4 + 11 x_2^{23} x_3^6 x_1 x_4^3\\
=&\ 12 x_1^7 x_2 x_3^5 - 18 x_1^3 x_4 + 11 x_1 x_2^{23} x_3^6 x_4^3\end{align*} Note that $\tau \circ \sigma = (1\ 3\ 4\ 2)$. Then
\begin{align*}&\ (\tau \circ \sigma) \cdot p\\
=&\ 12 x_3^5 x_1^7 x_2 - 18 x_1^3 x_4 + 11 x_3^6 x_1 x_4^3 x_2^{23}\\
=&\ 12 x_1^7 x_2 x_3^5 - 18 x_1^3 x_4 + 11 x_1 x_2^{23} x_3^6 x_4^3\end{align*} Note that $\sigma \circ \tau = (1\ 3\ 2\ 4)$. Then
\begin{align*}&\ (\sigma \circ \tau) \cdot p\\
=&\ 12 x_4^5 x_2^7 x_3 - 18 x_2^3 x_1 + 11 x_4^6 x_2 x_1^3 x_3^{23}\\
=&\ 12 x_2^7 x_3 x_4^5 - 18 x_1 x_2^3 + 11 x_1^3 x_2 x_3^{23} x_4^6\end{align*} (2) Let $p = \sum_{i \in I} a_i \prod_{j=1}^4 x_j^{r_{i,j}}$. We have \begin{align*}1 \cdot p =&\ 1 \cdot \sum_{i \in I} a_i \prod_{j=1}^4 x_j^{r_{i,j}} = \sum_{i \in I} a_i \prod_{j=1}^4 x_{\mathsf{id}(j)}^{r_{i,j}} \\=&\ \sum_{i \in I} a_i \prod_{j=1}^4 x_j^{r_{i,j}} = p.\end{align*} Now let $\sigma$, $\tau \in S_4$. Then \begin{align*}\sigma \cdot (\tau \cdot p) =&\ \sigma \cdot (\tau \cdot \sum_{i \in I} a_i \prod_{j=1}^4 x_j^{r_{i,j}}) = \sigma \cdot \sum_{i \in I} a_i \prod_{j=1}^4 x_{\tau{j}}^{r_{i,j}} \\=&\ \sum_{i \in I} a_i \prod_{j=1}^4 x_{\sigma(\tau(j))}^{r_{i,j}} = \sum_{i \in I} a_i \prod_{j=1}^4 x_{(\sigma \circ \tau)(j)}^{r_{i,j}}\\ =&\ (\sigma \circ \tau) \cdot \sum_{i \in I} a_i \prod_{j=1}^4 x_j^{r_{i,j}}.\end{align*} Thus we have a group action of $S_4$ on $R$.

(3) The permutations that stabilize $x_4$ are precisely those permutations that fix 4; i.e., $$\{ 1, (1\ 2), (1\ 3), (2\ 3), (1\ 2\ 3), (1\ 3\ 2) \}.$$ Clearly then this stabilizer is isomorphic to $S_3$.

(4) Any permutation which stabilizes $x_1 + x_2$ must either fix or swap 1 and 2; how it acts on 3 and 4 doesn’t matter. There are 4 such permutations: $A = \{ 1, (1\ 2), (3\ 4), (1\ 2)(3\ 4) \}$. We can fill in a multiplication table for this subset of $S_4$:

1(1 2)(3 4)(1 2)(3 4)
11(1 2)(3 4)(1 2)(3 4)
(1 2)(1 2)1(1 2)(3 4)(3 4)
(3 4)(3 4)(1 2)(3 4)1(1 2)
(1 2)(3 4)(1 2)(3 4)(3 4)(1 2)1

We can see from the table that $A$ is closed under multiplication and inversion, so it is a subgroup of $S_4$ of order 4. Moreover, we know from a previous exercise that because the multiplication table is symmetric, $A$ is abelian.

(5) Any permutation which stabilizes $x_1 x_2 + x_3 x_4$ must (independently) (1) swap or fix 1 and 2, (2) swap or fix 3 and 4, and (3) swap or fix the indices of the two terms. This gives $2^3 = 8$ permutations as follows: $$A = \{ 1, (1\ 2), (3\ 4), (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3), (1\ 3\ 2\ 4), (1\ 4\ 2\ 3) \}.$$ Now let $S = \{ r, s \}$ and definea mapping $\overline{\varphi} : S \rightarrow A$ by $\overline{\varphi})(r) = (1\ 3\ 2\ 4)$ and $\overline{\varphi}(s) = (1\ 2)$. It is easy to see that $\overline{\varphi}(r)^4 = \overline{\varphi}(s)^2 = 1$, that $\overline{\varphi}(r) \overline{\varphi}(s) = \overline{\varphi}(s) \overline{\varphi}(r)^{-1}$, that $\overline{\varphi}(s) \notin \langle \overline{\varphi}(r) \rangle$, and that $|\overline{\varphi}(r)| = 4$. By a lemma to a previous problem, then, $\overline{\varphi}$ extends to an injective group homomorphism $\varphi : D_8 \rightarrow A$. Since $|D_8| = |A|$, $\varphi$ is an isomorphism.

(6) As before, we can see that any permutation which stabilizes $(x_1 + x_2)(x_3 + x_4)$ must (independently) (1) swap or fix 1 and 2, (2) swap or fix 3 and 4, and (3) swap or fix the indices of the two factors. This gives the same set of permutations as in the previous part.


Linearity

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