Show that G is isomorphic to C4

Question

I know that the order of G is 4, but it is not the Klein-four group because not all of the elements have order of 2. Since G has elements of order 4 it is cyclic. I also know by definition that the order of C4 is 4.

But I don't know where to go from here to show they are isomorphic. Any hints or tips would be great, thank you!

Answer 1

Hint. The elements $i^0, i^1, i^2, i^3$ are pairwise distinct.

Answer 2

Map a generator of $G$, say $i$ or $-i$, onto a generator of $C_4=(\mathbf Z/4\mathbf Z, +)$, say $1$ or $3+4\mathbf Z$.

Another approach: the elements of $G$ can be written as $\Bigl\{\mathrm e^{\tfrac{ik\pi}2}\mid k\in\mathbf Z\Bigr\}$. Show that for each element in $G$, the corresponding values of $k$ are congruent modulo $4$ and deduce that the map $\mathbf Z/4\mathbf Z\longrightarrow G$, $x=k+\mathbf Z\longmapsto\mathrm e^{\tfrac{ik\pi}2}$ is well-defined and check it is an isomorphism.

Answer 3

You are basically done.

Up to isomorphism there are only two groups of order 4, $C_4$ and $C_2 \times C_2$. In $C_2 \times C_2$, any element is a self inverse, which is not the case for your group. So it must be isomorphic to $C_4$.

Also, what you did is equally correct: showing that it's cyclic would be sufficient to show it's isomorphic to $C_4$.