What is the subgroup of $C^{*}$ generated by $2i?$ I know $i$ generates $\{1, -1, i, -i\}$, however in this case I get $2i, -4, -8i, 16, 32i, -64, -128i, 256,\dots$ Is it infinite? Why is the group generated by $2i$ so different from the one generated by $i$?   
 A: Let $G$ be a (multiplicative) subgroup of $\mathbb{C}^*$ generated by $2i$. If you define $\mathbb{Z}\to G$ by $n\mapsto (2i)^n$ you will quickly see that it is an isomorphism (with addition in $\mathbb{Z}$ and multiplication in $G$) because the kernel is trivial. So yes, it is infinite cyclic.
More generally if $G$ is any group and $g\in G$ then the mapping $\mathbb{Z}\to G$ given by $n\mapsto g^n$ is an epimorphism onto $<g>$. Thus by the first isomorphism theorem the image $<g>$ is isomprhic to $\mathbb{Z}/N$ for some subgroup $N$. In this case $N=\{0\}$ is trivial.
So the reason it is so different from the $i$ case is simply because $i^4=1$ while there is no $n\in\mathbb{N}$, $n\neq 0$ such that $(2i)^n=1$ (that's exactly why the kernel is trivial).
A: Yes, it is infinite, as you get more and more powers of $2$, as you discovered yourself. 
Why is it different from the group generated by $i$? I guess one answer could be, that $i$ has finite order, therefore generates only a finite group, while $2i$ has infinite order, 'because' $2$ has infinite order.  
