Multiplicative Group of Reals is Normal Subgroup of Complex

Theorem
Let $$\left({\R^*, \times}\right)$$ be the Multiplicative Group of Real Numbers.

Let $$\left({\C^*, \times}\right)$$ be the Multiplicative Group of Complex Numbers.

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

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

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

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

Then we note that $\left({\C^*, \times}\right)$ is abelian.

The result follows from All Subgroups of Abelian Group are Normal.