41 Group Actions and Permutation Representations

#todo/school : Do problem 6 📅 2025-02-20 ✅ 2025-02-19

6

Question

As in 22 Centralizers and Normalizers, Stabilizers and Kernels (exercise 12, which we didn't do) is described below:
Let $R$ be the set of all polynomials with integer coefficients in the independent variables $x_{1}, x_{2}, x_{3}, x_{4}$ , so the members of $R$ are finite sums of elements of the form $a x_{1}^{r_{1}} x_{2}^{r_{2}} x_{3}^{r_{3}} x_{4}^{r_{4}}$ , where $a$ is any integer and $r_{1}, \dots, r_{4}$ are non-negative integers. For example:

x = 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}

is a typical element of $R$ . Each $σ \in S_{4}$ (see Symmetric Group & Permutations) gives a permutation of ${x_{1}, \dots, x_{4}}$ by defining $σ \cdot x_{i} = x_{σ (i)}$ . This may be extended to a map from $R$ to $R$ by defining:

σ \cdot p (x_{1}, x_{2}, x_{3}, x_{4}) = p (x_{σ (1)}, x_{σ (2)}, x_{σ (3)}, x_{σ (4)})

For example, if $σ = (1 2) (3 4)$ and $p (x_{1}, \dots, x_{4})$ is the polynomial $x$ defined above, then

\begin{aligned} σ \cdot p (x_{1}, x_{2}, x_{3}, x_{4}) & = 12 x_{2}^{5} x_{1}^{7} x_{3} - 18 x_{1}^{3} x_{4} + 11 x_{2}^{6} x_{1} x_{4}^{3} x_{3}^{23} \\ = 12 x_{1}^{7} x_{2}^{5} x_{3} - 18 x_{1}^{3} x_{4} + 11 x_{1} x_{2}^{6} x_{3}^{23} x_{4}^{3} \end{aligned}

Now for the exercise:
a. Find the polynomials in the Orbit of $S_{4}$ on $R$ containing $x_{1} + x_{2}$ (the Stabilizers of this polynomial has Order (Groups) 4), which is ${σ \cdot r | r \in {(x_{1} + x_{2})}}$
b. Find the polynomials in the Orbit of $S_{4}$ on $R$ containing $x_{1} x_{2} + x_{3} x_{4}$ (the Stabilizers of this polynomial have order 8)
c. Find the polynomials in the orbit of $S_{4}$ on $R$ containing $(x_{1} + x_{2}) (x_{3} + x_{4})$

Proof

a. Notice if $G = S_{4}$ then $| G | = | S_{4} | = 4! = 24$ . Then using the Stabilizers definition:

\begin{aligned} G_{x_{1} + x_{2}} & = {σ \in S_{4} | σ \cdot (x_{1} + x_{2}) = (x_{1} + x_{2})} \\ = {σ \in S_{4} | x_{σ (1)} + x_{σ (2)} = x_{1} + x_{2}} \\ = {1, (1 2), (3 4), (1 2) (3 4)} \end{aligned}

where the last step comes from considering the $σ$ 's such that $1 \mapsto 1 \lor 2$ and $2 \mapsto 1 \lor 2$ and everything else stays the same. Now using the Orbit Stabilizer Theorem then:

O (x_{1} + x_{2}) ≅ \frac{G}{G_{x_{1} + x_{2}}}

So using Lagrange's Divisibility Theorem of Order of Subgroups then:

| O_{x_{1} + x_{2}} | = \frac{| G |}{| G_{x_{1} + x_{2}} |} = \frac{24}{4} = 6

so we expect an order 6 subgroup. By the Isomorphism we can make left cosets to see what polynomials we expect in the orbit.

After seeing the pattern, it's clear we can choose one representative from each to get the orbit:

\begin{aligned} O_{x_{1} + x_{2}} & = {1 \cdot (x_{1} + x_{2}), (1 3) \cdot (x_{1} + x_{2}), \dots} \\ = {x_{2} + x_{1}, x_{2} + x_{3}, x_{2} + x_{4}, x_{1} + x_{3}, x_{1} + x_{4}, x_{3} + x_{4}} \end{aligned}

b. Similar to (a) the size of the orbit is:

| O_{x_{1} x_{2} + x_{3} x_{4}} | = \frac{| G |}{| G_{x_{1} x_{2} + x_{3} x_{4}} |} = \frac{24}{8} = 3

so only 3 representatives are needed to be found. Here we had:

\begin{aligned} G_{x_{1} x_{2} + x_{3} x_{4}} & = {σ \in S_{4} | σ \cdot (x_{1} x_{2} + x_{3} x_{4}) = x_{1} x_{2} + x_{3} x_{4}} \\ = {σ | x_{σ (1)} x_{σ (2)} + x_{σ (3) σ (4)} = x_{1} x_{2} + x_{3} x_{4}} \end{aligned}

considering the cases:

$n$	$σ (n) =$	Possible Combinations of this
1	1	$(3 4)$ , $1$ ,
	2	$(1 2)$ , $(1 2) (3 4)$
	3	$(1 3) (2 4)$ , $(1 3 2 4)$
	4	$(1 4) (2 3)$ , $(1 4 2 3)$
2	2	contained above
	3	contained above
	4	contained above
3	3	...
	4	...
4	4	...
which is the right amount. Taking left-cosets gives:

\begin{aligned} O_{x_{1} x_{2} + x_{3} x_{4}} & = {(3 4) \cdot (x_{1} x_{2} + x_{3} x_{4}), 1 \cdot (x_{1} x_{2} + x_{3} x_{4}), \dots} \\ = {x_{1} x_{2} + x_{3} x_{4}, x_{1} x_{2} + x_{4} x_{3}, x_{2} x_{1} + x_{3} x_{4}, x_{2} x_{1} + x_{4} x_{3} \\ \dots, x_{3} x_{4} + x_{1} x_{2}, x_{3} x_{4} + x_{2} x_{1}, x_{4} x_{3} + x_{2} x_{1}, x_{4} x_{3} + x_{1} x_{2}} \end{aligned}

c. Let's find $G_{(x_{1} + x_{2}) (x_{3} + x_{4})}$ here! Notice that this is:

\begin{aligned} G_{(x_{1} + x_{2}) (x_{3} + x_{4})} & = {σ | (x_{σ (1)} + x_{σ (2)}) (x_{σ (3)} + x_{σ (4)}) = (x_{1} + x_{2}) (x_{3} + x_{4})} \end{aligned}

But this is actually just the same as (b), with multiplication and addition swapped. As such then the same permutations should work:

G_{(x_{1} + x_{2}) (x_{3} + x_{4})} = {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)}

Thus then our orbits can be dictates by applying these permutations:

\begin{aligned} O & = {(x_{1} + x_{2}) (x_{3} + x_{4}), (x_{2} + x_{1}) (x_{3} + x_{4}) \\ \dots (x_{1} + x_{2}) (x_{4} + x_{3}), (x_{2} + x_{1}) (x_{4} + x_{3}) \\ \dots (x_{3} + x_{4}) (x_{1} + x_{2}), (x_{4} + x_{3}) (x_{2} + x_{1}) \\ \dots (x_{3} + x_{4}) (x_{2} + x_{1}), (x_{4} + x_{3}) (x_{1} + x_{2})} \end{aligned}

☐