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This question is from Dummit and Foote:

Let $R$ be the set of all polynomials with integer coefficients in the independent variables $x_1, x_2, x_3, x_4$, i.e., the members of $R$ are finite sums of elements of the form $ax_1^{r_1}x_2^{r_2}x_3^{r_3}x_4^{r_4}$, where $a$ is any integer and $r_1,.. .,r_4$ are non negative integers. Each $\sigma\in S_4$ gives a permutation of $\{x_1,..., x_4\}$ by defining $\sigma\cdot x_i=x_{\sigma(i)}$. This may be extended to a map from $R$ to $R$ by defining $\sigma\cdot p(x_1,\dots,x_4)=p(x_{\sigma(1)},\dots,x_{\sigma(4)})$ for all $p(x_1,\dots,x_4)\in R$.
Exhibit all permutations in $S_4$ that stabilize the element $x_1+x_2$.

I don't understand what does last line mean. (I do know what 'stabilize' means in context 'action'.)

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  • $\begingroup$ Where is the problem? The equation $\sigma\cdot p(x_1,\dots,x_4)=p(x_{\sigma(1)},\dots,x_{\sigma(4)})$ defines an action of $S_4$ on $R$. $\endgroup$ – Arnaud D. Mar 22 '18 at 16:51
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You have a natural action of $S_{4}$ on $R = \mathbb{Z}[x_{1},x_{2},x_{3},x_{4}]$, where your permutations permute the indeterminates of your polynomial ring. Some polynomials will be fixed/stabilized by certain permutations of the variables, meaning a polynomial $p$ and a permutation $\varphi$ satisfy $\varphi(p) = p$. It is asking you to find the permutations $\varphi \in S_{4}$ such that $\varphi(x_{1}+x_{2}) = x_{1}+x_{2}$.

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You are looking for $\sigma\in S_4$ such that $\sigma.(x_1+x_2)=x_1+x_2$. For example, this holds when $\sigma=(12)$ since $(12).(x_1+x_2)=x_2+x_1=x_1+x_2$.

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"$\sigma$ stabilizes $x_1+x_2$'' means that $\sigma\cdot (x_1+x_2)=x_1+x_2$.

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The order of the stabilizer is known to be the order of the group divided by the order of the orbit. The order of the orbit of $x_1+x_2$ is $6$, as it can go to any $x_i+x_j$ with $i\not = j$ (but $x_i+x_j=x_j+x_i$)...

Thus the stabilizer has order $4$.

We get that it equals $\{(12),(34),(12)(34),e\}$.

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  • 1
    $\begingroup$ The asker didn’t request a complete solution to the exercise, but was trying to understand what it meant. $\endgroup$ – Davislor Mar 22 '18 at 22:12

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