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Showing that dihedral 4 is isomorphic to subgroup of permutations

  1. Feb 22, 2017 #1
    1. The problem statement, all variables and given/known data
    D4 acts on the vertices of the square. Labeling them counterclockwise
    starting from the top left as 1, 2, 3, 4, find the corresponding homomorphism
    to S4.

    2. Relevant equations


    3. The attempt at a solution

    I am not completely sure what the question is asking. It's pretty clear to see which elements of D4 would correspond to which permutations of S4, so am I being asked to just list which elements of D4 correspond to which permutations of S4? How would I prove the the resulting map is a homomorphism? It seems like it would be tedious to check all values to see if the homomorphism property is always satisfied.
     
  2. jcsd
  3. Feb 22, 2017 #2

    fresh_42

    Staff: Mentor

    That's how I see it. In the end, it's all about how you define ##D_4##. E.g. it can be defined as the group generated by two elements ##r\, , \,s## with the relations ##r^2=s^2=(rs)^4=1## where it's not immediately obvious how the representation by a monomorphism ##\tau## to ##\mathcal{Sym}(4)## works.
    Yes, that's true. Of course you could simply look it up on Wikipedia, or you try to find an argument, why the structure is preserved, i.e. why ##\tau (a\cdot b) = \tau (a) \cdot \tau(b)## holds and why ##\tau## is injective.
    The elements of ##D_4## are rotations by ##90°## and reflections along both middle axis of a square. So you have to find an argument, why two such operations in succession ##a\cdot b## map (via ##\tau##) to the same corresponding succession of permutations ##\tau (a) \cdot \tau(b)\,.##
     
  4. Feb 22, 2017 #3
    Could the idea of group action help me out here?
     
  5. Feb 22, 2017 #4

    fresh_42

    Staff: Mentor

    Yes. Just make sure that it doesn't become a terminological overkill here. You don't need the entire apparatus of group operations.

    Only ask what the elements in these groups are and how multiplication is defined on them. Is there any difference between the two sides of ##\tau : D_4 \hookrightarrow \mathcal{Sym}(4)## ? Or what is ##\tau^{-1}(\tau (a) \circ \tau (b))## ? The homomorphy lies already in the concept itself, if you define ##D_4## as a group of transformations instead of as a group with generators and relations. That's why I asked about the definition of ##D_4##. This determines the amount of work that has to be done. If you meant this by group action, then the answer to your question is yes.
     
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