Additive Group of Reals is Subgroup of Complex

Theorem

Let $\struct {\R, +}$ be the additive group of real numbers.

Let $\struct {\C, +}$ be the additive group of complex numbers.

Then $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.

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 $x + y = \paren {x_1 + y_1} + \paren {0 + 0} i$ 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, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.

$\blacksquare$

Sources

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