# A Counter example for direct summand

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

## 2 Answers

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