Quotient groups of the Gaussian integers of the form $\mathbb{Z}[i]/ c\mathbb{Z}[i]$ I was thinking about groups of the form $\mathbb{Z}[i] / c\mathbb{Z}[i]$ (which I will call $\mathbb{Z}[i]_c$ from now on) where $c$ is a Gaussian integer and addition is always the group operation.
Just as how we can model $\mathbb{Z}_n$ with sets of the form $\{r+qn | q\in \mathbb{Z} \}$ and $r$ held constant, I tried doing the same for Gaussian integers. This led to the creation of this Desmos graph, the details of which are explained in the graph.
Using the above graph, I noticed that $\mathbb{Z}[i]_{1+i} \cong \mathbb{Z}_2$. There’s a lot of ways to prove that $\mathbb{Z}[i]_{1+i}$ has two elements, and we can see that $f(a+bi)=a+b \mod 2$ works as an isomorphism between the two groups.
I also considered $\mathbb{Z}[i]_2$, and unlike $\mathbb{Z}[i]_{1+i}$, it doesn’t seem to be isomorphic to any cyclic group. It has four elements ${0,1,i,1+i}$, with the following addition table:
$$
\begin{array}{ c | c | c | c | c |}
+ & 0 & 1 & i & 1+i \\
\hline
0 & 0 & 1 & i & 1+i \\
\hline
1 & 1 & 0 & 1+i & i \\
\hline
i & i & 1 + i & 0 & 1 \\
\hline
1+i & 1+i & i & 1 & 0 \\
\hline
\end{array}
$$
Which is isomorphic to $\mathbb{Z}_2^2$.
After experimenting with a couple other of these groups, it seems like $|\mathbb{Z}[i]_{a+bi}|=a^2+b^2$, but I don’t know how to prove this. I mainly looked into $\mathbb{Z}[i]_{2+2i}$ and $\mathbb{Z}[i]_{1+2i}$. I think the second is isomorphic to $\mathbb{Z}_5$, and I have no idea what’s going with the first.
My question(s) are as follows:

*

*Am I correct in my conjecture that $|\mathbb{Z}[i]_{a+bi}|=a^2+b^2$


*Are all of these isomorphic to a cyclic group or some direct product of cyclic groups? If yes, what’s the pattern behind them?
 A: The rule is that the group is isomorphic to:
$$\mathbb Z/\frac{a^2+b^2}{\gcd(a,b)}\mathbb Z\times \mathbb Z/\gcd(a,b)\mathbb Z.$$
So you are correct when $\gcd(a,b)=1.$
Here's an approach that doesn't use ring quotients.
As additive groups, your subgroup for $a+bi$ is generated by $a+bi$ and $-b+ai=i(a+bi).$
If you can show that $\mathbb Z[i]/\langle a+bi,-b+ai\rangle$ always has $a^2+b^2$ elements, you get a long way towards your result.
There is a clever geometric way to show this, if you know Pick's formula.
The key to this proof is that $a+bi$ and $-b+ai$ are perpendicular and the same distance from $0,$ so the subgroup divides the plane into squares.
All elements of $\mathbb Z[i]$ are congruent to some elements, either inside the square bounded by the four points: $$0,a+bi,-b+ai,(a-b)+(b+a)i,$$ or one of the $2\gcd(a,b)-1$ boundary points on the line segments $0,a+bi,$ and $0,-b+ai.$  If $I$ is the number of interior points, then number of elements in your quotient is $I+2\gcd(a,b)-1.$
Using Pick's formula, the area is equal $\frac{1}{2}4\gcd(a,b)+I-1,$ where $4\gcd(a,b)$ is the number of points on the boundary of the square, and $I$ is the number of interior points. So the area is equal to the number of elements in the quotient group.
But the area is $a^2+b^2,$ since each side is length $\sqrt{a^2+b^2}.$

Claim: If $\gcd(a,b)=1,$ the smallest positive integer $n$ such that $(a+bi)(c+di)=n$ with $c,d\in\mathbb Z,$ is $n=a^2+b^2.$
Proof: Note that the imaginary part of $(a+bi)(c+di)$ is $ad+bc.$ For that to be zero, $ad=-bc.$ But $a\mid -bc$ Means $a\mid c.$ Likewise, $b\mid d.$ Solving, we get $d=an,c=-bn$ for some integer $n,$ and $(a+bi)(c+di)=(a^2 +b^2)n.$

This in turns means that no two of $0,1,\dots,a^2+b^2-1$ differ by $a+bi.$ So this is a complete set of representatives of our quotient group.

If $d=\gcd(a,b)>1,$ the smallest positive $n$ which is a multiple of $a+bi$ is $n=\frac{a^2+b^2}{d}.$ This shows that no element of $\mathbb Z[i]$ generates the whole quotient group, because $nz=0$ for all elements of the quotient group.
So the group is not cyclic if $d>1.$
Proving it is the product above is a little harder.
