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Homework Help: Find isomorphism

  1. Jul 28, 2011 #1
    Is what I did all I need to do? Is there anything else I need to prove?

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110728_175901.jpg?t=1311905852 [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110728_175917.jpg?t=1311905865 [Broken]
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 28, 2011 #2
    Hi Shackleford! :smile:

    You're probably not going to like this, but you're not done yet. You have yet to show that this is an isomorphism.

    Your function f is an isomorphism if it is bijective (this is trivial) and if


    for all x and y. You have to show that this holds for your groups.

    For example, you'll need to show that


    and so on for every element. This is a lot of work, but maybe you can find some shortcuts that could reduce the calculations a bit? Or maybe you won't bother with checking all of these things :smile:
  4. Jul 28, 2011 #3
    Am I correct in making the transformation by assigning the matrices e, a, b, and ab respectively? If so, then I already showed ab = [ ] [ ] = ba. I need to do a2, too?
  5. Jul 28, 2011 #4
    Yes, you're matrices are assigned perfectly!! :smile: The assignment you wrote down is an homomorphism, but you just need to prove it.
  6. Jul 29, 2011 #5
    One thing to consider is that both groups are abelian. This means that you don't need to check ab and ba. You just need to check one. So I think there's really only 6 mappings you need to check and it is better to do it in a certain order, I think.
  7. Jul 29, 2011 #6
    I need to check six mappings?

    The properties for isomorphism are one-to-one correspondence and, generally, if φ(a*b) = φ(a) x φ(b).

    Clearly it has the one-to-one correspondence, and I've already shown φ(ab) = φ(a)φ(b). As mentioned previously, I suppose I need to do a2 and b2.
  8. Jul 29, 2011 #7
    Don't you need to show it for a*ab b*ab and ab*ab. These will be trivial, but I don't see how you can get around it.
  9. Jul 29, 2011 #8
    I suppose you're right, since the set is so small.
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