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Show that fiber products exist in the category of abelian groups. In fact, If $X, Y$ are abelian groups with homomorphisms $f: X \to Z$ and $g: Y \to Z$ show that $X \times_z Y$ is the set of all pairs $(x, y)$ with $x \in X$ and $y \in Y$ such that $f(x) = g(y)$.

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What have you tried so far? Can you prove that $X \times_Z Y$ as defined in the last sentence has maps to $X$ and $Y$ so that the appropriate square from the definition commutes? That's the start to your problem; after that, you can deal with the universal property. – Michael Joyce Apr 20 '12 at 2:27
Since you are told one group that will work as a fiber product, what you really need to do is show that the given set/group has the universal property property of the fiber product. – Arturo Magidin Apr 20 '12 at 2:55
This is exercise 50(a) in chapter 1 of Lang's Algebra, verbatim. – Brett Frankel Apr 20 '12 at 3:06

Hint: You definitively know that $\mathbf{Ab}$ has products and equalizers (kernels). Use this to show that it has fiber products by constructing the fiber product construction as the equalizer of some product of maps.

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