Remember that $d$ is a greatest common divisor of $a$ and $b$ if and only if:
- $d|a$ and $d|b$; and
- If $c|a$ and $c|b$, then $c|d$.
Let $a$ and $b$ be integers, and let $d$ be their greatest common divisor as integers. Then we know that $d$ satisfies the two properties above, where "divides" means "divides in $\mathbb{Z}$; and $c$ in point 2 is an arbitrary integer.
The thing that makes this very easy is that in the integers, we can express a gcd as a linear combination: we know that there exist integers $\alpha$ and $\beta$ such that $d=\alpha a + \beta b$.
Thus, if $\mathbf{x}$ is a Gaussian integer that divides $a$ and divides $b$ (in the Gaussian integers), then it also divides $\alpha a$, it also divides $\beta b$, and hence also divides $\alpha a + \beta b$. Thus, if $\mathbf{x}$ divides $a$ and divides $b$ in the Gaussian integers, then it divides $d$ in the Gaussian integers.
Thus, $d$ is a greatest common divisor for $a$ and $b$ in the Gaussian integers, since it satisfies the two requirements to be a greatest common divisor:
- $d|a$ and $d|b$ (in the Gaussian integers); and
- If $c$ is a Gaussian integer that divides both $a$ and $b$, then it divides $d$ (in the Gaussian integers).
Thus, $d$ is a greatest common divisor for $a$ and $b$ in $\mathbb{Z}[i]$.
No need to muck about with norms or with products.
P.S. Note that in the integers we can talk about "the" greatest common divisor because, even though every pair of numbers (except for $0$ and $0$) has two greatest common divisors ($d$ and $-d$), there is a natural way to "choose" one of them. In the Gaussian integer each pair of numbers (again, except $0$ and $0$) has four greatest common divisors (if $d$ is one, then so are $-d$, $id$, and $=id$); there is no universally accepted way of deciding which one would be "the" greatest common divisor, so generally you should talk about a greatest common divisor rather than the greatest common divisor.
Best Answer
To prove that the product of gcd and lcm of $r,s$ is $rs$, let $m=\text{lcm}(r,s)$ and $d=\text{gcd}(r,s)$, then show $rs|md$ and $md|rs$.
Hint:
To show $rs|md$, let $r=ud, s=vd$, then $rs=(uvd)d$ and use the fact that $m$ divides any common multiplier of $r,s$.
To show $md|rs$, use the fact that $d=ir+js$ for some integers $i,j$.
$$\{e, a^s, a^{2s}, ..., a^{(d-1)s}\}$$
is the set of elements that satisfies the equation and the elements are distinct.