Group homomorphisms from $\mathbb{C}^\ast$ to $ \mathbb{Z}$ I want to find $Hom_{\mathtt{Grp}}(\mathbb{C}^\ast,\mathbb{Z})$, where $\mathbb{C}^\ast$ is the multiplicative group, and $\mathbb{Z}$ is additive.
$\mathbb{C}$ is the additive group of complex numbers. We have the following map: 
$\large{\mathbb{C} \xrightarrow{exp} \mathbb{C}^\ast \xrightarrow{?} \mathbb{Z}}$
where the fiber of $exp$ is $\mathbb{Z}$
And I don't know if this can help, any hint?
 A: A group homomorphism $\phi:\mathbb{C}^* \to \mathbb{Z}$ must be trivial.
Let $\omega \in \mathbb{C}^*$ and $n \in \mathbb N$. Then there is $\theta \in \mathbb{C}^*$ such that $\omega=\theta^n$ and so $\phi(\omega)= n \phi(\theta)$.
Therefore, $\phi(\omega)$ is a multiple of $n$ for all $n \in \mathbb N$. This can only happen if $\phi(\omega)=0$.
In short, $\mathbb{C}^*$ is a divisible group but $\mathbb{Z}$ is not. (The image of $\phi$ is either trivial or isomorphic to $\mathbb{Z}$.)
A: This isn't a full answer, but I suspect that this Hom group may be the trivial group. Suppose $\phi:\mathbb{C}^* \to \mathbb{Z}$ is a group homomorphism. We know that $\phi(1) = 0$ since $1$ is the identity. Then
$$
 \phi(-1)^2 = \phi(1) = 0 \implies 2 \phi(-1) = 0 \implies \phi(-1) = 0
$$
By a similar argument, any complex number $e^{2\pi i/k}$ where $k \in \mathbb{Z}$ should go to zero, since
$$
\phi(e^{2\pi i/k})^{k} = \phi(2\pi i) = \phi(1) = 0 \implies k \phi(e^{2\pi i/k}) = 0 \implies \phi(e^{2\pi i/k}) = 0
$$
So we have a dense subset of the unit circle that all must get sent to zero. I don't quite know how to use this, but it seems likely to me that this will force $\phi$ to send everything to zero.
