# Gaussian Integers form Subgroup of Complex Numbers under Addition

## Theorem

The set of Gaussian integers $\Z \sqbrk i$, under the operation of complex addition, forms a subgroup of the set of additive group of complex numbers $\struct {\C, +}$.

## Proof

We will use the One-Step Subgroup Test.

This is valid, as the Gaussian integers are a subset of the complex numbers.

We note that $\Z \sqbrk i$ is not empty, as (for example) $0 + 0 i \in \Z \sqbrk i$.

Let $a + b i, c + d i \in \Z \sqbrk i$.

Then we have $-\paren {c + d i} = -c - d i$, and so:

\(\displaystyle \paren {a + b i} + \paren {-\paren {c + d i} }\) | \(=\) | \(\displaystyle \paren {a + b i} + \paren {- c - d i}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a + \paren {-c} } + \paren {b + \paren {-d} } i\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a - c} + \paren {b - d} i\) |

We have that $a, b, c, d \in \Z$ and $\Z$ is an integral domain.

Therefore by definition $\Z$ is a ring.

So it follows that $a - c \in \Z$ and $b - d \in \Z$, and hence $\paren {a - c} + \paren {b - d} i \in \Z \sqbrk i$.

So by the One-Step Subgroup Test, $\Z \sqbrk i$ is a subgroup of $\struct {\C, +}$.

$\blacksquare$

## Sources

- 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): Chapter $\text{II}$: Groups: Direct Products