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

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.

    1. \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
    2. \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
    3. Note that \tau \circ \sigma = (1\ 3\ 4\ 2). Then
      (\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
    4. Note that \sigma \circ \tau = (1\ 3\ 2\ 4). Then
      (\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
  1. Let p = \sum_{i \in I} a_i \prod_{j=1}^4 x_j^{r_{i,j}}. We have 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. Now let \sigma, \tau \in S_4. Then \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}}. Thus we have a group action of S_4 on R.
  2. 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.
  3. 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)
    1 1 (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.

  4. 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.
  5. 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.
Post a comment or leave a trackback: Trackback URL.

Comments

  • Gobi Ree  On October 19, 2011 at 3:38 am

    In 6, the answer seems to be based on the fact that R=\mathbb{Z}[x_1,x_2,x_3,x_4] is an UFD.

    • nbloomf  On October 19, 2011 at 10:03 am

      The UFD-hood of R probably does play a role- since the action of G 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.

  • Gobi Ree  On October 19, 2011 at 3:48 am

    Also are the subscripts correct? I think that all the x_i^{r_{i,j}} (in this problem and also in the next problem) should be changed to x_{i,j}^{r_{i,j}}.

    • nbloomf  On October 19, 2011 at 10:04 am

      The subscripts on x are indeed wrong; it should read x_j. Thanks!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: