Are finite abelian groups isomorphic to k-character groups for arbitrary alg.closed k?

Question

Let $G$ be a finite abelian group, $k$ an algebraically closed field with group of units $k^{\times}$, and let $Ch_{k}(G)$ be the set of $k$-characters of $G$ whose elements are multiplicative homomorphisms of the form $\chi: G \to k^{\times}$ which forms a group under pointwise multiplication. On page 62 of Etingof et al. "Introduction to Representation theory" it is shown that $Ch_{\mathbb{C}}(G)$ is non-canonically isomorphic to $G$. Is this also true for an arbitrary algebraically closed field $k$?

Answer 1

Without constraint on the characteristic $p$ of the field $k$, a finite abelian group is not isomorphic to its $k$-character group. To be explicit, let $p > 0$ and $G \cong \mathbf Z/p\mathbf Z$. Because the field $k$ does not have 'enough' $p$-th roots of unity (as $x^p - 1 = (x - 1)^p$ holds in $k$), we get $$|\operatorname{Hom}(G, k^\times)| = |\{\, x \in k : x^p = 1 \,\}| = 1.$$ So it cannot be isomorphic to $G$.