the nonzero elements of Z3[i] form an abelian group of order 8 under multiplication. Is it isomorphic to Z8??

Question

$\mathbb{Z}/3\mathbb{Z}[i]$ is an integral domain, so its characteristic is a prime number.

But, in order to prove that it is isomorphic to $\mathbb{Z}_8$, we have to show that $\mathbb{Z}_3[i]$ has an element of order $8$ with respect to multiplication.

Am I proceeding in the right manner?

Answer 1

Yes, compute the orders of the $8$ elements. If one has order $8$, you are done.

In fact, you do not need to check them all, since $\mathbb{Z}_8$ has $4$ generators. So checking at most $5$ objects will do the job.

Compute the orders of $1$, $-1$, $i$, $-i$, and $1+i$. The first four are easy to deal with. They all have order $\le 4$. As to $1+i$, its square is $-i$, and squaring again gives $-1$, so the order is $8$.

Of course we could have checked $1+i$ first!

Answer 2

Another way of approach $\mathbb{Z}/3\mathbb{Z}[i] \cong \mathbb{F}_3[X]/(X^2+1)\cong \mathbb{F}_9$. And $\mathbb{F}_9^{*}\cong C_8$, a cyclic group of order $8$.