# Multiplicative Group of Rationals is Subgroup of Reals

## Theorem

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

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

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

## Proof

From the definition of real numbers, it follows that $\Q_{\ne 0}$ is a subset of $\R_{\ne 0}$.

As $\struct {\R_{\ne 0}, \times}$ is a group, and $\struct {\Q_{\ne 0}, \times}$ is a group, it follows from the definition of subgroup that $\struct {\Q_{\ne 0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$.

As $\struct {\R_{\ne 0}, \times}$ is abelian, it follows from Subgroup of Abelian Group is Normal that $\struct {\Q_{\ne 0}, \times}$ is normal in $\struct {\R_{\ne 0}, \times}$.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 36$: Subgroups: Simple illustrations: $(1)$ - 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 $4.3$