Additive Group of Reals is Normal Subgroup of Complex

Theorem
Let $\left({\R, +}\right)$ be the additive group of real numbers.

Let $\left({\C, +}\right)$ be the additive group of complex numbers.

Then $\left({\R, +}\right)$ is a normal subgroup of $\left({\C, +}\right)$.

Proof
Let $x, y \in \C$ such that $x = x_1 + 0 i, y = y_1 + 0 i$.

As $x$ and $y$ are wholly real, we have that $x, y \in \R$.

Then $x + y = \left({x_1 + y_1}\right) + \left({0 + 0}\right)i$ which is also wholly real.

Also, the inverse of $x$ is $-x = -x_1 + 0 i$ which is also wholly real.

Thus by the Two-Step Subgroup Test, $\left({\R, +}\right)$ is a subgroup of $\left({\C, +}\right)$.

Then from Complex Numbers under Addition form Abelian Group, $\left({\C, +}\right)$ is abelian.

The result follows from Subgroup of Abelian Group is Normal.