Gaussian Integers form Subgroup of Complex Numbers under Addition

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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