Proof involving homomorphism between $\Bbb Z^n$ and an abelian group G

Question

Let $(e_1, ..., e_n)$ be the standard $\Bbb Z$-basis of $\Bbb Z^n$. Let $x_1,..., x_n$ be elements of an abelian group $G$.

I want to prove that there exists a homomorphism $f$ $:$ $\Bbb Z^n$ $\rightarrow$ $G$ such that $f$$(e_i)$ $=$ $x_i$ for all $i$.

Please can anyone lend a hand here?

Answer 1

Hint: $(a_1, \dots, a_n)= a_1 e_1 + \dots + a_n e_n$.

Answer 2

The only thing you need to show is that it's possible to map any element $x = \sum_{k=1}^n a_k x_k $ into G in a way that makes the whole map a group homomorphism. Now there's only one way, given by (where I write $G $ additively)

$f (x) = \sum_{k=1}^n a_k f (x_k) $.

All you need to check is that this is a well-defined homomorphism, which is fairly obvious from the construction.