Gaussian Integers form Subgroup of Complex Numbers under Addition

From ProofWiki
Jump to navigation Jump to search


The set of Gaussian integers $\Z \left[{i}\right]$, under the operation of complex addition, forms a subgroup of the set of additive group of complex numbers $\left({\C, +}\right)$.


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 $\Z \left[{i}\right]$ is not empty, as (for example) $0 + 0 i \in \Z \left[{i}\right]$.

Let $a + b i, c + d i \in \Z \left[{i}\right]$.

Then we have $- \left({c + d i}\right) = -c - d i$, and so:

\(\displaystyle \left({a + b i}\right) + \left({- \left({c + d i}\right)}\right)\) \(=\) \(\displaystyle \left({a + b i}\right) + \left({- c - d i}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({a + \left({-c}\right)}\right) + \left({b + \left({-d}\right)}\right) i\)
\(\displaystyle \) \(=\) \(\displaystyle \left({a - c}\right) + \left({b - d}\right)i\)

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 $\left({a - c}\right) + \left({b - d}\right)i \in \Z \left[{i}\right]$.

So by the One-Step Subgroup Test, $\Z \left[{i}\right]$ is a subgroup of $\left({\C, +}\right)$.