$\operatorname{Hom}(G,\Bbb C\setminus \{0\})$ non- isomorphic to $\operatorname{Hom}(G, \Bbb T)$?

Question

Do you have an example of an abelian group $G$ for which $\operatorname{Hom}(G,\Bbb C\setminus \{0\})$ is not isomorphic to $\operatorname{Hom}(G, \Bbb T)$?

$\Bbb C$ is the complex plane and $\Bbb T $ is the unit circle group in the complex plane. $\operatorname{Hom}$ denotes homomorphism group.

Answer 1

Considered as abstract groups, $\mathbb{C}\setminus\{0\}\cong\mathbb{T}$ (both are isomorphic to the direct sum of $\mathbb{Q}/\mathbb{Z}$ and a vector space over $\mathbb{Q}$ of continuum dimension). So $\operatorname{Hom}(G,\mathbb{C}\setminus\{0\})\cong\operatorname{Hom}(G,\mathbb{T})$ as abstract groups for every $G$.