To prove the $(\Bbb R , +)$ is isomorphic to $(\Bbb C , +)$ without using metric space?

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Firstly, I want to show $(\Bbb R , +)$ is isomorphic to $\Bbb R^2$ (under component wise addition).

In this I used the mapping $\phi_1$ : $\Bbb R^2 \rightarrow \Bbb R$

Where, $\phi_1(a,b) = 2^a3^b$

And then , show $\Bbb R^2$ is isomorphic to $(\Bbb C , +)$.

In this I used the mapping

$\phi_2$ : $\Bbb R^2 \rightarrow \Bbb C$

Where, $\phi_2(a,b) = a+ib$

And then, use the transitive property of isomorphic groups to show $(\Bbb R , +)$ is isomorphic to $(\Bbb C , +)$

Now ,my doubts are that, whether the mappings i used are correct or not. Does this mappings makes the two groups isomorphic?

Please, correct me wherever I am wrong?

Edit: I see that $\phi_1$ is not bijective. So , their is no way to prove ,$\phi_1$ : $\Bbb R^2 \rightarrow \Bbb R$ is an isomorphism by just defining a mapping or without using the concept of metric space!