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The exercise is: Let $G$ be an abelian group containing a subset $X = \left\{ x_1 , \ldots , x_n \right\}$ and suppose for every abelian group $H$ and every function $\gamma : X \rightarrow H$ there exists a unique homomorphism $\tilde{\gamma} : G \rightarrow H $ such that $\tilde{\gamma} \left(x_j\right) = \gamma \left(x_j \right)$ for every $j$. Show that $G$ is free abelian of rank $n$.

Now, it should be sufficient to show that $X$ is the basis, but how would I go about doing this? I tried using $X$ to generate $\sum_{x \in X} \mathbb{Z}$, a direct sum, and finding an isomorphism back to $G$, but without success. I'm sure that $\tilde{\gamma} \left(x_j\right) = \gamma \left(x_j \right)$ is quite important, though I don't see how it fits into the puzzle at all.

Any help at all is much appreciated. Thank you.

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You can pretty much do what you described. Notice $\sum_{x \in X} \cong \mathbb{Z}^n$ (In fact, they're pretty much the same.)

Show that $\mathbb{Z}^n$ with $X'=\{e_1,\ldots,e_n\}$ has the same universal property: Every map into a group $H$ given only on the base $X'$ can be uniquely extended to a group homomorphism $\mathbb{Z}^n \to H$. Then try to show that any two such groups with this universal property are isomorphic, or maybe just do that very special case, to conclude $G \cong \mathbb{Z}^n$.

This you can indeed do by constructing a homomorphism $G \to \mathbb{Z}^n$ using the universal property with the obvious choice of an image of $X$. Then you can use the universal property of $\mathbb{Z}^n$ to find a morphism back.

Look at the compositions of these morphisms $G \to \mathbb{Z}^n \to G$ and $\mathbb{Z}^n \to G \to \mathbb{Z}^n$ and the images of $X$ and $X'$. Again you can work with the universal property to show that the morphisms you defined are inverse.

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    Dude, no problem. I'm just procrastinating anyway. If not for you, I probably just would have watched some stupid youtube videos.2012-10-10