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i have learn that intersection of pure subgroup of a group G is not necessarily pure. Can someone show me an example when such a case exists?

I'm aware that if G is torsion-free, then intersection of pure subgroup of G are necessarily pure. So the example above must involve for which the group G is not torsion-free.

Any idea? thanks

edit: The group G here we are talking about is abelian, of course.

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up vote 3 down vote accepted

This is exercise 10.33(i) in Rotman's "An Introduction to Group Theory". You can take $G=\mathbb Z_2 \times \mathbb Z_8$. Take $A$ the subgroup generated by (0,1) and $B$ the subgroup generated by (1,1). Then $A$ and $B$ are pure, but their intersection will be the subgroup generated by (0,2), which is not pure in $G$.

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One can just verify the definition, but a result of Kulakoff may put the example in context. A bounded pure subgroup, like A and B, must be a direct summand. Since A∩B is not a direct summand, but is bounded, it cannot be a pure subgroup. In other words, intersections of direct summands need not be direct summands. "Pure" is the "limit" of "direct summand," so in some sense this is really the fundamental example. – Jack Schmidt Jan 24 '11 at 18:11
Does the subgroup $A$ looks like $<(1,0)>=\{(0,0),(1,0)\}$ and $B$ looks like $<(1,1)>=\{(1,1),(0,2),(1,3),(0,4),(1,5),(0,6),(1,7),(0,0)\}.$ Then the intersection of their subgroup will not be generated by (0,2) or do you mean that $A$ is generated by $(0,1)$? – Seoral Jan 25 '11 at 5:31
The example I worked out took two cyclic subgroups of maximal order (so A=(0,1) as you say). They must be direct summands by Kulakoff. Their intersection is contained in G^p, so cannot be a nonzero direct summand (not being pure). – Jack Schmidt Jan 25 '11 at 15:06
Seoral: I corrected the misprint: A is generated by (0,1). – mathreader Jan 26 '11 at 11:00
Jack: You mention that their intersection is contained in $G^p$... what is $G^p$? Sorry for the late revert. I just come back from chinese new year. Cheers. – Seoral Feb 9 '11 at 7:04

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