Let be the set of all polynomials with integers coefficients in the independent variables ; i.e., elements of are formal sums for some finite set , integers , and nonnegative integers .
Each permutes the by . We extend this action to a mapping by . That is, permutes the indices of the variables.
 Let , and let and . Compute , , , and .
 Prove that this set of mappings gives a group action of on .
 Exhibit all permutations in that stabilize and prove that they form a subgroup isomorphic to .
 Exhibit all permutations in that stabilize the element and prove that they form an abelian group of order 4.
 Exhibit all permutations in that stabilize the element and prove that they form a subgroup isomorphic to .
 Show that the permutations in that stabilize the element are exactly those found in the previous part.

 Note that . Then
 Note that . Then
 Let . We have . Now let . Then . Thus we have a group action of on .
 The permutations that stabilize are precisely those permutations that fix ; i.e., . Clearly then this stabilizer is isomorphic to .
 Any permutation which stabilizes must either fix or swap 1 and 2; how it acts on 3 and 4 doesn’t matter. There are 4 such permutations: . We can fill in a multiplication table for this subset of :
We can see from the table that is closed under multiplication and inversion, so it is a subgroup of of order 4. Moreover, we know from a previous exercise that because the multiplication table is symmetric, is abelian.
 Any permutation which stabilizes 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 permutations as follows: . Now let and definea mapping by and . It is easy to see that , that , that , and that . By a lemma to a previous problem, then, extends to an injective group homomorphism . Since , is an isomorphism.
 As before, we can see that any permutation which stabilizes 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.
Comments
In 6, the answer seems to be based on the fact that is an UFD.
The UFDhood of probably does play a role since the action of sends irreducibles to irreducibles it must permute the (essentially unique) irreducible factors of any given polynomial. There’s probably something more to be said here.
Also are the subscripts correct? I think that all the (in this problem and also in the next problem) should be changed to .
The subscripts on are indeed wrong; it should read . Thanks!