Why is $\{1,-1,i,-i\}$ isomorphic to the cyclic group $C_4$?

Question

I'm reading the Wu-ki Tung's book Group Theory in Physics, where it is stated that the group $\{\pm 1, \pm i\}$ under the usual multiplication is isomorphic to the cyclic group of order $4$, $C_4$. If I understood correctly, the multiplication table for $C_4$ is $$ \begin{array}{c|cccc} & I & A & B & C \\ \hline I & I & A & B & C \\ A & A & B & C & I \\ B & B & C & I & A \\ C & C & I & A & B \\ \end{array} $$ while the table for $\{1, -1, i, -i\} \equiv \{I,A,B,C\}$ is $$ \begin{array}{c|cccc} & I & A & B & C \\ \hline I & I & A & B & C \\ A & A & I & C & B \\ B & B & C & A & I \\ C & C & B & I & A \\ \end{array} $$

Both tables are different and cannot be made equal by just relabeling the elements. Why are these groups isomorphic to each other?

Answer 1

The cyclic group of order 4 is the abelian group $\{1,a,a^2,a^3\}$, and the other group is $\{1,i,i^2,i^3\}$ so the isomorphism is quite easy to see. In your question you've put $-1$ as the second element, which is probably what confused you.

Answer 2

The order 2 element is $-1$, and the order 4 elements are $\pm i$. So if you let $B = -1$, $A = i$ and $C = -i$ in the second table you'll find the same table as above.

Answer 3

Isomoprphisms send identity to identity and preserve elements' orders. So, defined $G:=\{1,-1,i,-i\}$, $C_4:=\{1,a,a^2,a^3\}$, and $\varphi\colon G\to C_4$ as a candidate isomorphism, necessarily: \begin{alignat}{1} \varphi(1) &:= 1 \\ \varphi(-1) &:= a^2 \\ \end{alignat} As for $\pm i$, both of order $4$, in order to get a bijection we are left with: \begin{alignat}{1} \varphi(i) &:= a \\ \varphi(-i) &:= a^3 \\ \end{alignat} or: \begin{alignat}{1} \varphi(i) &:= a^3 \\ \varphi(-i) &:= a \\ \end{alignat} and both options lead indeed to an isomorphism. So, we have actually two of them.