Showing that a subgroup of Sym(4) is isomorphic to D_8

  • Thread starter Thread starter Mr Davis 97
  • Start date Start date
  • Tags Tags
    Subgroup
Click For Summary
SUMMARY

The subgroup of the symmetric group ##S_4## that stabilizes the polynomial ##x_1x_2 + x_3x_4## is identified as ##\{1,(12),(34),(12)(34),(13)(24),(14)(23),(1324),(1423)\}##. This subgroup is proven to be isomorphic to the dihedral group ##D_8##, which represents the symmetries of a square. The correspondence between the elements of the subgroup and the symmetries of the square is established through a detailed analysis of the permutations. The discussion emphasizes the computational complexity of exhibiting this correspondence in detail.

PREREQUISITES
  • Understanding of group theory concepts, specifically symmetric groups and dihedral groups.
  • Familiarity with polynomial functions and their properties in multiple variables.
  • Knowledge of group actions and stabilizers in the context of abstract algebra.
  • Basic combinatorial techniques for listing permutations and analyzing their properties.
NEXT STEPS
  • Study the properties of dihedral groups, focusing on ##D_8## and its applications in symmetry.
  • Learn about group actions and stabilizers in greater depth, particularly in relation to polynomial functions.
  • Explore the relationship between permutations and geometric transformations, especially in the context of symmetry.
  • Investigate computational methods for establishing isomorphisms between groups, including practical examples.
USEFUL FOR

Mathematics students, particularly those studying abstract algebra, group theory, and polynomial functions, as well as educators seeking to enhance their understanding of group actions and symmetries.

Mr Davis 97
Messages
1,461
Reaction score
44

Homework Statement


Let ##R## be the set of all polynomials with integer coefficients in the independent variables ##x_1, x_2, x_3, x_4##.
##S_4## acts on ##R## by the group action ##\sigma \cdot p(x_1,x_2,x_3,x_4) = p(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)},x_{\sigma(4)})##. 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 the dihedral group of order 8.

Homework Equations

The Attempt at a Solution


So, by listing the elements of ##S_4## and seeing how they act on ##x_1x_2+x_3x_4##, I found that the stabilizer of ##x_1x_2+x_3x_4## is ##\{1,(12),(34),(12)(34),(13)(24),(14)(23),(1324),(1423) \}##.

My problem is showing that this is isomorphic to the dihedral group of order 8. I think the correspondence is clear if we label the vertices of a square and consider the permutations as symmetries of the square, but I feel that exhibiting this 1-1 correspondence in that much detail is too computationally intensive and that there should be an easier way.
 
Physics news on Phys.org
Mr Davis 97 said:
So, by listing the elements of ##S_4## and seeing how they act on ##x_1x_2+x_3x_4##, I found that the stabilizer of ##x_1x_2+x_3x_4## is ##\{1,(12),(34),(12)(34),(13)(24),(14)(23),(1324),(1423) \}##.

My problem is showing that this is isomorphic to the dihedral group of order 8. I think the correspondence is clear if we label the vertices of a square and consider the permutations as symmetries of the square, but I feel that exhibiting this 1-1 correspondence in that much detail is too computationally intensive and that there should be an easier way.
What is ##D_8\,##? And no, I'm not asking about the index, nor do I want to know that it is the dihedral group of order eight. Without this information, I'd say pin a screenshot of this Wiki page and you're done: https://de.wikipedia.org/wiki/Diedergruppe#Permutations-Darstellung
Just observe, that you have an odd numbering of the square: ##(12)## and ##(34)## are the diagonals and not neighboring points.
 
  • Like
Likes   Reactions: Mr Davis 97

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 21 ·
Replies
21
Views
1K
Replies
28
Views
6K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
3K
  • · Replies 27 ·
Replies
27
Views
11K
  • · Replies 32 ·
2
Replies
32
Views
3K
Replies
2
Views
2K