Are $\mathbb{Z} \times \mathbb{Z}_2$ and $ \mathbb{Z}$ isomorphic?

Question

Are $\mathbb{Z} \times \mathbb{Z}_2$ and $ \mathbb{Z}$ isomorphic ?

My attempt : I think No because $\mathbb{Z} \times \mathbb{Z}_2$ is not cyclic but $ \mathbb{Z}$ is cyclic

Edit : Suppose there is an isomorphism $f : \mathbb{Z} \times \mathbb{Z_2} \to \mathbb{Z} $.Then $\ker f =\{f \in\mathbb{Z} \times \mathbb{Z}_2 : f(0,0)=0 \}$

This implies that there exist an element $f(0,1) \in \mathbb{Z}$ such that $f(0,1)^2=f(0,1) + f(0,1)=f(0+0,1+1)=f(0,2)=f(0,0)=0$

i,e there exist an an element of order $2$ in $\mathbb{Z}$

This leads to contradiction because if we take the $f(0,1)=k \in \mathbb{Z} $ then $f(0,1)^2=k +k=2k =0$ .This implies $(0,1) \in \ker f$.

Answer 1

Your attempt is a good start. Now, prove that $\mathbb Z\times \mathbb Z_2$ is not cyclic. It should be relatively simple.

Alternatively, you could also:

• note that one of the two groups has a finite subgroup, and the other one does not.
• Note that specifically, $\mathbb Z_2$ is a subgroup of one of the groups and not the other
• Note that all nonzero elements of $\mathbb Z$ have infinite order, while $\mathbb Z\times \mathbb Z_2$ has an element of finite order. This argument can actually also be used to prove that $\mathbb Z\times\mathbb Z_2$ is not cyclic.

All of these are (somewhat related) arguments showing that the two groups are not isomorphic.