Cancellation of Groups in Internal Direct Products

  • Thread starter Thread starter Oster
  • Start date Start date
  • Tags Tags
    Groups Internal
Oster
Messages
84
Reaction score
0
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?
 
Physics news on Phys.org
Oster said:
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 \phi : G \times H \to G \times K can be written in terms of maps \theta: G \to G and \psi : H \to K as
\phi(g,h) = (\theta(g),\psi(h))
Now work out the conditions \theta and \psi must satisfy for \phi to be an isomorphism.
 
Last edited:
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?
 
Oster said:
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 \phi(g,h) = (\theta(g,h),\psi(g,h)) for \theta : G \times H \to G and \psi : G \times H \to K. The idea is then to look at \phi(1,h).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top