Question: The non-zero elements of $\mathbb{Z}_{3}\left [ i \right ]$ form an abelian group of order 8 under multiplication. Is it isomorphic to $\mathbb{Z}_{8},\mathbb{Z}_{4}\bigoplus \mathbb{Z}_{4}$ or $\mathbb{Z}_{2}\bigoplus \mathbb{Z}_{2}\bigoplus\mathbb{Z}_{2}$.
I think the stratagem is to find the number of orders for each elements in $\mathbb{Z}_{3}\left [ i \right ]$ and for all the other groups. Since Isomorphism demands bijection, if the number of elements in the other groups equals to that of the number of elements of certain order in $\mathbb{Z}_{3}\left [ i \right ]$, we are done.
Note: $\mathbb{Z}_{3}\left [ i \right ]=\left \{1,2,i,1+i,2+i,2i,1+2i,2+2i \right \}$
There is a duplicate thread on this but I felt the solution odd. Nonetheless I have left a comment on that thread.
Any help is appreciated. Thanks in advance.
Consider $\langle 1+i\rangle$:
$$(1+i) \mapsto (2i) \mapsto (1+2i) \mapsto (2) \mapsto (2+2i) \mapsto (i) \mapsto (2+i) \mapsto (1)$$
Hence $\Bbb Z_3[i]$ is cyclic, and is isomorphic to $\Bbb Z_8$.