## Does A/B' = A/B imply B' = B?

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?

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 Recognitions: Gold Member Science Advisor Staff Emeritus 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.
 Recognitions: Gold Member Science Advisor Staff Emeritus 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.