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I know it's a theorem that every exact sequence of abelian groups splits, but I'm guessing this problem is supposed to be solvable by smaller guns. Given an exact sequence of abelian groups $0 \to A \to B \to \mathbb{Z}^{n} \to 0$, we wish to show that $B \cong A \times \mathbb{Z}^{n}$.

I really know only the basics about short exact sequences and splitting homomorphisms, so I'm not sure how to proceed here without just applying the big theorem.

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    $\begingroup$ That's not a theorem, though. There's an exact sequence $0 \to \mathbb{Z}_2 \to \mathbb{Z}_4 \to \mathbb{Z}_2 \to 0$ that can't split. $\endgroup$
    – Randall
    Aug 21, 2017 at 0:33

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$B/A \cong \mathbf{Z}^n$ means that $B/A$ is generated freely by some elements $b_1 + A, \dots, b_n + A$. Use this to get a map $\mathbf{Z}^n \to B$.

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