So the multiplicative group of complex numbers, $\mathbb{C}^*$, is divisible as an abelian group. Why is the multiplicative group of real numbers, $\mathbb{R}^*$, not divisible as an abelian group? What is the difference?
My lecturer remarked this as an explanation : "$\mathbb{R}^*$ contains $-1$ which is not divisible by $2$".
I don't really understand this, because you can clearly write $-1 = 2(-\frac{1}{2})$
"Divisible" means with respect to the $\Bbb Z$-module structure of the abelian group $(G,*)$, id est the action $$n\odot z=\begin{cases}\underbrace{z*z*\cdots*z}_{n\text{ times}}&\text{if }n\ge 0\\ \underbrace{z^{-1}*z^{-1}*\cdots*z^{-1}}_{(-n)\text{ times}}&\text{if }n< 0\end{cases}$$ Since the operation, in this cases, is multiplication, the action of $\Bbb Z$ is $n\odot z:=z^n$.
Concretely, here being divisible by $2$ means having a square root.