For which $G$, we have $Hom(G, \mathbb Z/p)\cong(\mathbb Z/p)^n$? Let $G$ be a finitely generated abelian group and $Hom(G, \mathbb Z/p)\cong(\mathbb Z/p)^n$. What can we say about $G$? Any suggestion? Thanks
 A: By the structure theorem on finite type abelian groups, $G$ is isomorphic to a direct sum (where the $p_i$ are prime numbers, not necessarily distinct):
$$G \simeq \mathbb Z^r \oplus \bigoplus_{i=1}^m \mathbb Z / p_i^{a_i}$$
And therefore $\mathrm{Hom}(G, \mathbb Z/p) \simeq \mathrm{Hom}(\mathbb Z, \mathbb Z/p)^r \oplus \bigoplus_{i=1}^m \mathrm{Hom}(\mathbb Z/p_i^{a_i}, \mathbb Z/p)$. It's also easy to prove that:
$$\mathrm{Hom}(H, \mathbb Z/p) \simeq
\begin{cases}
\mathbb Z/p & H = \mathbb Z \\
\mathbb Z/p & H = \mathbb Z/p^k \\
0 & H = \mathbb Z/q^k \wedge q \neq p
\end{cases}$$
The conclusion is that $r + |\{ i : p_i = p\}| = n$ (it's a necessary and sufficient condition).
A: In general, if $F$ is a field and $G$ is an abelian group, then, treating $F$ as an additive group, $\mathrm{Hom}(G,F)$ can be made a vector space over $F$ in a simple fashion.
Now, if $G$ is finitely generated, then show that the vector space must be finite-dimensional. (Hint: The value of any homomorphism from $G$ to $F$ is entirely determined by the values on the generators.)
This shows that for all finitely generated $G$ there exists an $n$ such that $\mathrm{Hom}(G,\mathbb Z/p)\cong (\mathbb Z/p)^n$. It doesn't tell you how to compute $n$.
