Are $\Bbb{R}$ and $\Bbb{Z}\times\Bbb{R}$ isomorphic as groups?

Question

If $\phi:\Bbb{R}\to\Bbb{Z}\times\Bbb{R}$ be group isomorphism. Then $\phi(1)=(n,x)$ for some $n\in\Bbb{Z},x\in\Bbb{R}$. Then $2n\phi(1/2n)=(n,x)\implies n=0\implies \phi(1)=(0,x)$

Now since $\phi$ is onto there is $y\in\Bbb{R}$ such that $\phi(y)=(1,0)$. It's easy to see that $y$ should be irrational.

I cannot proceed further, can anyone provide me an idea how to prove/disprove the statement?

Thanks for your help in advance.

Answer 1

Suppose $$\varphi\colon \mathbb{Z}\times\mathbb{R}\rightarrow \mathbb{R}$$ is an isomorphism.

Let $\varphi(1,0) = x\in \mathbb{R}$. By surjectivity there exists a $(a,b)\in \mathbb{Z}\times \mathbb{R}$ such that $\varphi(a,b) = \frac{x}{2}$. So we have: \begin{align*} \varphi(1,0) = x = 2\varphi(a,b) = \varphi(2a,2b) \end{align*} By injectivity we have $(1,0) = (2a,2b)$. So $2a = 1$, but $a\in \mathbb{Z}$ which is a contradiction.

Answer 2

$\mathbb Z$ is a homomorphic image of $\mathbb Z\times\mathbb R$; the projection map $(x,y)\mapsto x$ is a homomorphism.

$\mathbb Z$ is not a homomorphic image of $\mathbb R$, since the positive sentence $\forall x\exists y(y+y=x)$ is true in $\mathbb R$ but not in $\mathbb Z$.

Therefore $\mathbb Z\times\mathbb R$ is not isomorphic to $\mathbb R$, nor even a homomorphic image of $\mathbb R$.

Answer 3

$\mathbb R$ is a divisible group, wheras $\mathbb Z\times\mathbb R$ is not.