Module homomorphism on the algebraic closure of $\mathbb{Z}_p$ Let $k$ be the algebraic closure of $\mathbb{Z}_p$ and $k^*=k-\{0\}$ be a multiplicative group. For a finite abelian group (i.e. a finite $\mathbb{Z}$-module) $G$, compute the following: $$\text{Hom}_{\mathbb{Z}} (G,k^*) \otimes_{\mathbb{Z}} \mathbb{Z}_p$$
I think it suffices to assume that $G \cong \mathbb{Z}_{p^{a_1}}\oplus \cdots \oplus \mathbb{Z}_{p^{a_k}}$; since any other parts should be zero in a homomorphism from $G$ to $k^*$. But no other ideas.
 A: Writing $G \cong \mathbb Z/n_1 \oplus \cdots \oplus \mathbb Z/n_r$, we can indeed reduce the problem to cyclic groups, since $$\text{Hom}_{\mathbb Z}(G,k^{\times}) \otimes_{\mathbb Z} \mathbb Z/p \cong \bigoplus_{i=1}^r \text{Hom}_{\mathbb Z}(\mathbb Z/n_i,k^{\times}) \otimes_{\mathbb Z} \mathbb Z/p,$$ which you can read either way by distributivity of the tensor product.
Now, each $\text{Hom}_{\mathbb Z}(\mathbb Z/n,k^{\times}) \cong \{\zeta\in k^{\times} \mid \zeta^n = 1\} \subseteq k^{\times}$ is a finite subgroup (of course the containment needs $k$ algebraically closed). This is the set otherwise known as $\mu_n(k)$ of $n$-th roots of unity in $k$.
The extension $\mathbb F_p(\mu_n(k)) / \mathbb F_p$ is algebraic, hence finite, so $\mathbb F_p(\mu_n(k)) = \mathbb F_q$ for some power $q$ of $p$. Of course, $\mathbb F_q^{\times}$ is a cyclic group of order $q-1$, and the subgroup $\mu_n(k) \subseteq \mathbb F_q^{\times}$ has order $(n,q-1)$. Altogether, we see that $$\text{Hom}_{\mathbb Z}(\mathbb Z/n,k^{\times}) \otimes_{\mathbb Z} \mathbb Z/p \cong \mathbb Z/(n,q-1) \otimes_{\mathbb Z} \mathbb Z/p \cong \mathbb Z/(n,q-1,p) = 0.$$
I'm sure this can be simplified a bit, you really only need to know that $p\nmid (q-1)$ for any $q=p^s$.
