Multiplicative Group of Reals is Normal Subgroup of Complex

Theorem
Let $\left({\R_{\ne 0}, \times}\right)$ be the Multiplicative Group of Real Numbers.

Let $\left({\C_{\ne 0}, \times}\right)$ be the Multiplicative Group of Complex Numbers.

Then $\left({\R_{\ne 0}, \times}\right)$ is a normal subgroup of $\left({\C_{\ne 0}, \times}\right)$.

Proof
Let $x, y \in \C_{\ne 0}$ such that $x = x_1 + 0 i, y = y_1 + 0 i$.

As $x$ and $y$ are wholly real, we have that $x, y \in \R_{\ne 0}$.

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

Also, the inverse of $x$ is $\dfrac 1 x = \dfrac 1 {x_1} + 0 i$ which is also wholly real.

Thus by the Two-Step Subgroup Test, $\left({\R_{\ne 0}, \times}\right)$ is a subgroup of $\left({\C_{\ne 0}, \times}\right)$.

Then we note that $\left({\C_{\ne 0}, \times}\right)$ is abelian.

The result follows from Subgroup of Abelian Group is Normal.