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I know that the assertion above doesn't hold for general abelian groups (e.g. $\prod_{i=1}^\infty \mathbb{Z}$), but I just wanted a quick verification that it holds for finite abelian groups.

Edit: Sorry I meant quotient out by a group isomorphic to $A$, not necessarily $A\times\{0\}$

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    $\begingroup$ When you quotient by $A$, do you really mean the specific subgroup $A \oplus 0 \subset A \oplus B$, or just any subgroup isomorphic to $A$? In the former case it holds even for infinite groups. $\endgroup$ – arkeet Sep 28 '16 at 23:29
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If the $A$ you quotient by is $A \oplus \{0\}$, then this holds for both infinite and finite abelian groups. If you mean that there is another injective homomorphism $A \to A \oplus B$ by whose image you take the quotient, then it does not hold for finite abelian groups either: consider $C_2 \to (C_2 \oplus C_4)$ which takes $1$ to $(0,2)$; the quotient is $C_2 \oplus C_2$.

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