Stuck: Finding an Isomorphism for an Invertible Ring I'm stuck on a problem creating an isomorphism between rings. Specifically, let $\mathbb{Z}[\sqrt{7}] = R$.
Then for the invertible group $(R/3R)^\times$, I want to find an isomorphism to another group (i.e. of the form $(R/3R)^\times \cong \mathbb{Z}/a\mathbb{Z} \times \mathbb{Z}/b\mathbb{Z}$).
It's here that I'm stuck, as I'm pretty new to number theory (started a couple weeks ago), so any help would be appreciated. Cheers

Since you're taking the quotient by the ideal $3R$, the ring $\mathbb{Z}/3\mathbb{Z}$ comes into play here: namely the elements of $R/3R$ essentially have the form $a+b\sqrt{7}$ where $a,b\in \{0,1,2\}$. Multiplication in $R/3R$ is done by first multiplying in $R$, then taking $a,b$ modulo $3$. So let's think of these as pairs $(a,b)$.
Thus $R/3R$ has nine elements, so $(R/3R)^\times$ has at most eight elements (you have to throw out zero). Which nonzero elements are units? Since we're dealing with a finite ring, the elements are either units or zero divisors.
For example $2\sqrt{7}$ is a unit mod $3R$ because $(2\sqrt{7})^2=28=1$ mod $3$. It's easy to see that for $(a,b)=(1,0),(2,0),(0,1),(0,2)$, the element $a+b\sqrt{7}$ is a unit mod $3R$ (it's its own inverse!).
On the other hand, the elements $1+\sqrt{7}$ and $1+2\sqrt{7}$ have zero product mod $3$, so neither are units. The same is true for $2+\sqrt{7}$ and $2+2\sqrt{7}$.
Conclusion: there are four units mod $3$.
$$(R/3R)^\times=\{1,2,\sqrt{7},2\sqrt{7}\}.$$
Every element has order two. So
$$(R/3R)^\times \simeq \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$$
which is the Klein $4$-group.
