# Multiplicative Group of Reals is Subgroup of Complex

## Theorem

Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.

Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.

Then $\struct {\R_{\ne 0}, \times}$ is a normal subgroup of $\struct {\C_{\ne 0}, \times}$.

## 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, $\struct {\R_{\ne 0}, \times}$ is a subgroup of $\struct {\C_{\ne 0}, \times}$.

From Non-Zero Complex Numbers under Multiplication form Abelian Group, $\struct {\C_{\ne 0}, \times}$ is abelian.

The result follows from Subgroup of Abelian Group is Normal.

$\blacksquare$

## Sources

- 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Examples of groups $\text{(ii)}$ - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $4$: Subgroups: Example $4.3$