1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cancellation of Groups in Internal Direct Products!

  1. Dec 24, 2012 #1
    G, H, K are groups. G is finite. GxH is isomorphic to GxK. Prove H is isomorphic to K. Give an example to show that this does not hold when G is infinite.

    The counter example when G is infinite is Rx{0} and RxR (R - real numbers)
    I'm having trouble Proving the main part of the question. I have a hunch that the image of Gx{0} in GxK will be the direct product of a subgroup of G and a subgroup of K. Can someone help me?
     
  2. jcsd
  3. Dec 24, 2012 #2

    pasmith

    User Avatar
    Homework Helper

    The first observation is that, for an isomorphism to exist between finite groups, the orders of the groups must be equal. So immediately H and K are of the same order.

    Any isomorphism [itex]\phi : G \times H \to G \times K[/itex] can be written in terms of maps [itex]\theta: G \to G[/itex] and [itex]\psi : H \to K[/itex] as
    [tex]\phi(g,h) = (\theta(g),\psi(h))[/tex]
    Now work out the conditions [itex]\theta[/itex] and [itex]\psi[/itex] must satisfy for [itex]\phi[/itex] to be an isomorphism.
     
    Last edited: Dec 24, 2012
  4. Dec 24, 2012 #3
    H and K need not be finite. And I don't see why your second claim should hold. Can you explain a bit more please?
     
  5. Dec 24, 2012 #4

    pasmith

    User Avatar
    Homework Helper

    I should have [itex]\phi(g,h) = (\theta(g,h),\psi(g,h))[/itex] for [itex]\theta : G \times H \to G[/itex] and [itex]\psi : G \times H \to K[/itex]. The idea is then to look at [itex]\phi(1,h)[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Cancellation of Groups in Internal Direct Products!
Loading...