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All my attempts proving the following claim have been useless and it seems to be wrong, but can not find any counter example(s) for it. :)

"If U and N be two direct summands of an abelian group G such that N+U=G, then the intersection of N and U is a direct summand of G also".

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@Jonas: Yes. As you noted, G is assumed to be abelian and N+U is referred the subgroup generated by N and U. Thanks. – Nancy Rutkowskie Apr 28 '11 at 17:16
Thank you. Whoops, I just noticed the tag! I hope you don't mind if I add "abelian" to the question itself. – Jonas Meyer Apr 28 '11 at 17:53
up vote 7 down vote accepted

There are counterexamples. One is given in another question on this site, Example of Intersection of Pure Subgroup which is not Pure. As Jack Schmidt points out in a comment, an example is given by $G=\mathbb{Z}_2\oplus\mathbb{Z}_8$, $N=\langle (1,1)\rangle$, and $U=\langle (0,1)\rangle$. Then $G$ is the direct sum of $N$ and $\langle (1,4)\rangle$, and of $U$ and $\langle(1,0)\rangle$. The intersection is $\langle(0,2)\rangle\cong\mathbb{Z}_4$, which is not a direct summand of $G$ because $G$ is not isomorphic to $\mathbb{Z}_4\oplus\mathbb{Z}_4$ or to $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_4$. The group generated by $N$ and $U$ is $G$ because it contains $(0,1)$ and $(1,0)=(1,1)-(0,1)$.

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If I'm remembering correctly the direct summands should have a trivial intersection. The trivial group is normal as is the entire group $G$ (just noticed your groups are abelian so that's not a problem!). So if you like you could consider $G = \{e \} \oplus G$.

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Thanks for the advice. We don not have G=N⊕U here. There is no need for N and U to be direct summand to each other. I mean, no need to have G=N⊕U here. We have just G=<N,U>. :) – Nancy Rutkowskie Apr 28 '11 at 17:24

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