# Set of homomorphisms on a free abelian group is a free abelian group.

If $$G$$ is a free abelian group with rank $$n$$, I need to show that $${\rm Hom}(G,\mathbb{Z})$$, set of all homomorphisms is also free abelian group of rank $$n$$,

My work:

Since $$G$$ is free abelian group then $$G$$ can be generated by the set $$\{g_1,g_2, \cdots,g_n\}$$. Let $$f \in{\rm Hom}(G,\mathbb{Z})$$, then $$f(x)=\sum_{i=1}^n{c_if(g_i)}$$, where $$x=\sum_{i=1}^n{c_ig_i}$$, but this gives a generator set of $${\rm Im}(f)$$ and I need to get generators for $${\rm Hom}(G,\mathbb{Z})$$, so for arbitrary $$f$$ we have the same set of generators which is in $$\mathbb{Z}$$.

• Consider homomorphisms $\phi_{i}\in\operatorname{Hom}(G,\Bbb Z)$ for which $\phi_{i}(g_i)=1$ and $\phi_i(g_j)=0$ for $j\ne i$. (Such $\phi_i$ exist, are unique, etc.) Commented May 6, 2022 at 13:20
• @HagenvonEitzen, that's it. Thank you very much! Commented May 6, 2022 at 13:24

The idea is the same used to get a basis for the dual space of a vector space $$V$$ of finite dimension.
Let us define the homomorphisms $$g_i^*\colon G\to \mathbb{Z}$$ such that $$g_i^*(g_j):=\delta_{ij}$$. Then given $$f\in Hom(G,\mathbb{Z})$$ you have that
$$f=\sum_{g_i}f(g_i)\cdot g_i^*$$
Now suppose that $$\sum_{g_i}a_ig_i^*=0$$. Then $$\left(\sum_{g_i}a_ig_i^*\right)(g_j)=a_j=0$$ for each $$j$$. Hence $$Hom(G,\mathbb{Z})$$ is a free abelian group of rank $$n$$.