Cancellation of Groups in Internal Direct Products!

  • Thread starter Oster
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  • #1
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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?
 

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  • #2
pasmith
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G, H, K are groups. G is finite. GxH is isomorphic to GxK. Prove H is isomorphic to K.

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.
 
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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?
 
  • #4
pasmith
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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?

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].
 

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