Suppose $G$ and $H$ are two free abelian groups of countable but not finite rank. Is a non-trivial homomorphism $f: G \rightarrow H$ necessarily injective?

Intuition says yes, but I am just worried about what happens when the rank is no longer finite. I am not comfortable with groups that are not finitely generated, as they rarely arise.


The obvious realisation of a countably generated free Abelian group is $\bigoplus_{i = 1}^\infty \mathbb{Z} $. That is, integer sequences $(x_i)_{i=1}^\infty$ with finite support (finitely many terms non-zero).

We can make a homomorphism from this set to itself, $(x_i)_{i = 1}^\infty \mapsto (x_1, 0, 0, \dots)$, collapsing all except the first term to zero, and this is pretty far from injective.

  • $\begingroup$ Hm, yeah I should have thought of that. I guess even going a step further: what if it is non-trivial in each coordinate? In each coordinate, a homomorphism is determined by where a generator goes, so surely it must be injective in that case. Then my question is more along the lines of: are there different hierarchies of "countable?" $\endgroup$ – moron4ever Sep 28 '16 at 0:10
  • $\begingroup$ If it's non-trivial in each co-ordinate then you're right that the homomorphism must be injective. That's because the image will also be a countably-generated (but not finitely generated) free Abelian group, and all such groups are isomorphic. $\endgroup$ – Stanley Sep 28 '16 at 0:16
  • $\begingroup$ Thank you, that was what I was wondering. $\endgroup$ – moron4ever Sep 28 '16 at 3:10
  • $\begingroup$ I actually got the above comment wrong. It's true that if the image of every member of the generating set is nonzero (or even if the images of infinitely many generators is nonzero) then the image is isomorphic to every countably generated free Abelian group. But the homomorphism we invent might not be the isomorphism from the domain to the image. For example if you send two generators to the same element then the homomorphism is not injective. $\endgroup$ – Stanley Sep 28 '16 at 9:30
  • $\begingroup$ I would disagree with the above comment. We can sent /every/ generator to the same generator in the image. Now they're all nontrivial, but the image is still isomorphic to the integers. I need that the set of images of the generators is linearly independent over Z. This will guarantee that the image is isomorphic to every countably generated free Abelian group. (But not that the homomorphism is an isomorphism: consider the endomorphism mapping every x to x+x.) $\endgroup$ – Alex Meiburg Nov 26 '16 at 7:47

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