Can one always construct a square-root of a homomorphism to $\mathbb{C}^*$?

Question

L.S.,

Let $G$ a group and $\chi: G \rightarrow \mathbb{C}^*$ a morphism of groups. If we set $\delta: G \rightarrow \mathbb{C}$ by $\delta(g) = \sqrt{\chi(g)}$, where we choose one branch of $\mathbb{C}$, is it then true $\delta$ a morphism of groups for which $\delta^2 = \chi$?

I would say it is true, since $\mathbb{C}$ is algebraically closed, and the square root is a morphism from $\mathbb{C}$ to $\mathbb{C}$, but I'm afraid I'm missing something..

Thanks!

Answer 1

Suppose that $G=C_2$ is cyclic of order 2 with generator $x$. Define $\chi$ by $\chi(1)=1$, $\chi(x)=-1$. A reasonable choice for the square root would be $\sqrt{1}=1$, $\sqrt{-1}=i$, but then $$ \sqrt{\chi(1)}=1\neq -1=\sqrt{\chi(x)}\sqrt{\chi(x)}$$

If we try to fix this by choosing $\sqrt{1}=-1$, then $$ \sqrt{\chi(1)}=-1\neq 1=\sqrt{\chi(1)}\sqrt{\chi(1)} $$

So in either case $\sqrt{\chi}$ is not a homomorphism.