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Suppose that $G_1$ and $G_2$ are direct summands of torsion free abelian group $G$. Must $G_1 \cap G_2$ also be a direct summand?

It is true when $G$ is free, at least.

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1 Answer 1

Assume that $G$ is free. Since $G_1,G_2$ are direct sumands of $G$ and every subgroup of a free abelian group is also free, we get that $G/G_1, G/G_2$ are free. On the other side, the kernel of the canonical map $G\to G/G_1\times G/G_2$, $g\mapsto(g\pmod{G_1},g\pmod{G_2})$ is $G_1\cap G_2$, so $G/G_1\cap G_2$ is a subgroup of the free group $G/G_1\times G/G_2$, hence it is also free. This is enough to show that $G_1\cap G_2$ is a direct summand of $G$.

When $G$ is torsion free and $G/G_1,G/G_2$ are finitely generated, the result still holds true (by the same argument as above).

In general, I don't think the result is true: if $G_1,G_2$ are indecomposable and $G_1\cap G_2\neq \{0\}$, then $G_1\cap G_2$ is a direct summand of $G$ iff $G_1=G_2$. Unfortunately, at this moment I don't have such an example with $G_1\neq G_2$.

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Actually, your proof shows that $G_1 \cap G_2$ is a direct summand of the abelian group $G$ if $G/G_i\;(i=1,2)$ is free. –  tj_ Jan 27 '13 at 1:52
    
@tj_ You are right. –  user26857 Jan 27 '13 at 11:11

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