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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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Ignore the previous question - it was written before your edit showed up. I'll give that a try and report back with any issues. Thank you very much. –  JoeDub Oct 10 '12 at 7:32
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You mean $X \to \mathbb{Z}^n$ since $X \subseteq G$, right? And yes, that would be a neat choice. Mapping a base to base. I mean, you have bases $X = \{x_1, \ldots, x_n\} \subseteq G$ and $X' = \{e_1, \ldots, e_n\} \subseteq \mathbb{Z}^n$. If you were to construct $\gamma\colon X \to \mathbb{Z}^n$ and $\delta \colon X' \to G$ to eventually get inverse group homomorphisms $\tilde\gamma$ and $\tilde\delta$ – what could you possibly do? Edit: Ok, I'll check later. –  k.stm Oct 10 '12 at 7:38
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I'm not sure, what you mean. Strictly speaking, $\gamma \colon X \to \mathbb{Z}^n$ and $\delta \colon X' \to G$ are no group homomorphisms, because $X$ and $X'$ are no groups. You construct them to get – by the universal property – extensions to group homomorphisms $\tilde\gamma\colon G \to \mathbb{Z}^n$ and $\tilde\gamma\colon \mathbb{Z}^n \to G$ which are inverse and satisfy $\gamma (x_i) = \tilde\gamma (x_i)$ and $\delta (e_i) = \tilde\delta (e_i)$ $\forall i = 1, \ldots , n$. Edit: Yes, for $f = \tilde\gamma$ and $g = \tilde\delta$. –  k.stm Oct 10 '12 at 8:28
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But, you have not given a definition of $\gamma$ and $\delta$, because: which $z_i$ and which $g_j$ would you use? They haven't been defined yet. But, I'd leave out them anyway. ; - ) –  k.stm Oct 10 '12 at 8:32
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EDIT: YES, I meant $\tilde\delta$, sorry. However, I used the tilde to denote the extension, just adepting your notation from your question. You want to show that $f \circ g = \mathrm{id}_{\mathbb{Z}^n}$ and $g \circ f = \mathrm{id}_G$. (But I wouldn't use $g$ for the map while there is a group $G$ on the scene, but instead the tilde versions of $\gamma$ and $\delta$.) And no, it doesn't, but you don't need that. –  k.stm Oct 10 '12 at 8:44

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