What is the Isomorphism between Groups and its Implications?

[Boorglar]

Homework Statement

Prove or disprove the following assertion. Let $G$, $H$, and $K$ be groups. If $G × K \cong H × K$, then $G \cong H$.

Homework Equations

$G × H = \left\{ (g,h): g \in G, h \in H \right\}$

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:

$\left\{0\right\} × Z$ and $Z_{2} × Z$ with the isomorphism being:
$\phi( 0, 2n ) = ( 0, n )$ and $\phi( 0, 2n + 1 ) = ( 1, n )$. Then it is well-defined, one-to-one and onto. But it does not preserve the operation because $\phi( 0, 1 ) + \phi( 0, 1 ) = ( 1, 0 ) + ( 1, 0 ) = ( 0, 0 )$ while $\phi( (0,1) + (0,1) ) = \phi(0,2) = (0,1)$.

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 $\phi : G → H$ defined as $\phi( g ) = h$ where $(g,e_{K}) → (h, k')$ under the isomorphism between $G × K$ and $H × K$.

Then I proved $\phi$ 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 $g_{1} ≠ g_{2}$ then $(g_{1},e_{K}) → (h_{1}, k_{1})$ and $(g_{2}, e_{K} ) → (h_{2}, k_{2})$ where $h_{1}≠h_{2}$.

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 :(

 

[micromass]

The group $\mathbb{Z}_2\times \mathbb{Z}$ has an element of order $2$, the group $\{0\}\times \mathbb{Z}$ doesn't. So the groups are not isomorphic.

You're however correct that the statement is false. Take the following group as $K$

$$\mathbb{Z}^{\mathbb{N}} = \{(x_n)_n~\vert~x_n\in \mathbb{Z}\}$$

with the usual operations. Can you find the right $G$ and $H$?

 

[Boorglar]

Is that the set of all infinite sequences of integers?

 

[micromass]

Yes.

 

[Boorglar]

Ok yes! I finally got it:

Let G be {0} and let H be Z. Then define $\phi$ to be:

$\phi( (0, (a_{1}, a_{2}, ... ) ) ) = ( a_{1}, (a_{2}, a_{3}, ... ) )$ obtained by left-shifting the sequence. It is obviously well-defined. It is one-to-one because if two left-shifted sequences are equal, then adding a zero in front keeps them equal. And any sequence can have a zero added to it so the mapping is onto. The operation is preserved because:

$\phi( (0, (a_{1}+b_{1}, a_{2}+b_{2}, ... ) ) ) = ( a_{1}+b_{1}, (a_{2}+b_{2}, a_{3}+b_{3}, ... ) )$ $= ( a_{1}, (a_{2}, a_{3}, ... ) ) + ( b_{1}, (b_{2}, b_{3}, ... ) ) = \phi( ( 0, ( a_{1}, a_{2}, ... ) ) ) + \phi( 0, ( b_{1}, b_{2}, ... ) ) )$

Thanks for the help. I don't think I would have thought of that anytime soon :P

 

[Boorglar]

By the way, are you really 19? Because that's my age too but you seem so much more knowledgeable!