Index of subgroup of Real number in Complex Number

Question

Complex Number is group under addition. Real Number is its subgroup. It is the trivially normal subgroup. I wanted to find its index.
I thought one 2 dimension plane with X axis and Y axis. Its coset can be given in form like $(0,a)+(R,0)$ Where $a$ is any real Number.
This then says that is index is same as the cardinality of R.
Is I am right?

Answer 1

You are right, the index is $|\Bbb R|$ because that's how many cosets $\Bbb R$ has in $\Bbb C$. Geometrically, each coset is a horizontal line in the complex plane, and each horizontal line is a coset. Your argument is an exact algebraic analogue of that.

Answer 2

Another way to see this: as abelian additive groups $\mathbb{C}=\mathbb{R} \oplus \mathbb{R}$. Here $\mathbb{R} \cong \{(r,0): r \in \mathbb{R}\}$. Now define a map $\mathbb{C} \rightarrow \mathbb{R}$, by $z \mapsto Im(z)$. This is a surjective homomorphism, with kernel $\mathbb{R}$. Hence $\mathbb{C}/\mathbb{R} \cong \mathbb{R}$ as additive groups.