Is A/B' = A/B a Sufficient Condition for B' = B in Abelian Groups?

[eok20]

I'm probably missing something obvious, but suppose that B' < B < A are all abelian groups and that A/B is isomorphic to A/B'. Does it follow that B = B'? In the case of finite groups and vector spaces it is true by counting orders and dimensions but what about in general?

 

[Hurkyl]

It's true if the if the isomorphism is compatible with the projection maps.

That is, it's not enough that there be some random isomorphism between the groups; the projection A/B' --> A/B must be an isomorphism.


As is usually the case, think about infinite subsets of the integers, and use them to construct a counter-example. The first one I came up with is:

Let A be the free Abelian group generated by the symbols [n] for each integer n. Let B be the subgroup generated by the symbols [2n], and let B' be the subgroup generated by the symbols [4n].

Then A / B and A / B' are both free Abelian groups generated by a countably infinite number of elements; they are isomorphic.

 

[Hurkyl]

It strikes me that, in my example, we could let B' be finitely generated, or even be the zero group, so that it's not even isomorphic to B.