- #1
Boorglar
- 210
- 10
Homework Statement
Prove or disprove the following assertion. Let [itex]G[/itex], [itex]H[/itex], and [itex]K[/itex] be groups. If [itex]G × K \cong H × K[/itex], then [itex]G \cong H[/itex].
Homework Equations
[itex] G × H = \left\{ (g,h): g \in G, h \in H \right\} [/itex]
The Attempt at a Solution
I don't even know whether the statement is true or false... I tried looking in both directions but to no avail.
For a counterexample, I figured that at least one of the sets would have to be infinite, otherwise G and H would have the same number of elements and it would be harder for them not to be isomorphic. I considered the sets:
[itex] \left\{0\right\} × Z [/itex] and [itex] Z_{2} × Z [/itex] with the isomorphism being:
[itex] \phi( 0, 2n ) = ( 0, n ) [/itex] and [itex]\phi( 0, 2n + 1 ) = ( 1, n ) [/itex]. Then it is well-defined, one-to-one and onto. But it does not preserve the operation because [itex] \phi( 0, 1 ) + \phi( 0, 1 ) = ( 1, 0 ) + ( 1, 0 ) = ( 0, 0 ) [/itex] while [itex] \phi( (0,1) + (0,1) ) = \phi(0,2) = (0,1) [/itex].
I still think someone could tweak the definition a bit to make it an isomorphism, but of course I am not sure.
As for proving the theorem is true, I don't even know how to start. I considered the following mapping [itex] \phi : G → H [/itex] defined as [itex] \phi( g ) = h [/itex] where [itex] (g,e_{K}) → (h, k') [/itex] under the isomorphism between [itex] G × K [/itex] and [itex] H × K [/itex].
Then I proved [itex] \phi [/itex] is well-defined and preserves the operation. But I cannot prove that it is one-to-one, nor onto. The problem is that I have no reason to believe that if [itex]g_{1} ≠ g_{2} [/itex] then [itex] (g_{1},e_{K}) → (h_{1}, k_{1}) [/itex] and [itex] (g_{2}, e_{K} ) → (h_{2}, k_{2}) [/itex] where [itex] h_{1}≠h_{2} [/itex].
On the other hand, no other natural candidate for an isomorphism comes to mind, so I am completely stuck here. If someone could, not give me the whole answer, but just point to the right direction? Because I may be looking somewhere completely unrelated :(