# If $G, H, K$ are divisible abelian groups and $G \oplus H \cong G \oplus K$ then $H \cong K$

This is an exercise in Hungerford. But can somebody explain why is the following not a counter-example?

Let $G$ be the direct sum of $|\mathbb{R}|$ copies of $\mathbb{Q}$. Let $K$ be the direct sum of $|\mathbb{N}|$ copies of $\mathbb{Q}$. Then $G \oplus \mathbb{Q} \cong G \oplus K$ but $\mathbb{Q}$ is not isomorphic to $K$.

Indeed, suppose $f : \mathbb{Q} \rightarrow K$ is a $\mathbb{Z}$-module isomorphism. We may show that $f$ is a $\mathbb{Q}$-module isomorphism obtaining thus a contradiction. For any $v \in Q$ non-zero and a non-zero natural $b$ there is a unique $w$ such that $bw = v$, that is $w = (1/b)v$. We have $b[(1/b)v] = v$ so $b f((1/b) v) = f(v)$. The same uniqueness argument applies in $K$ since it is torsion-free, so $f((1/b) v) = (1/b) f(v)$. Hence $f$ is a $\mathbb{Q}$-module isomorphism.

• Can you produce an actual isomorphism $\;G\oplus\Bbb Q\to G\oplus K\;$ ? I'm afraid that what may be clear setwise is not that straightforward groupwise... – DonAntonio Nov 2 '13 at 20:06
• @DonAntonio: Same number of $\mathbb{Q}$s. – Martin Brandenburg Nov 2 '13 at 20:08
• Can you cite the specific exercise so others can look it up? – anon Nov 2 '13 at 20:09
• It's exercise 12, part b, in page 199 – Pedro Nov 2 '13 at 20:10
• I don't see that clear, @Martin: an element in $\;G\oplus\Bbb Q\;$ is of the form $\;(a,q)\;,\;a\in G\,,\,q\in\Bbb Q\;$ , whereas an element in $\;G\oplus K\;$ is of the form $\;(a,k)\;,\;a\in G\,,\,k\in K\;$ . The second coordinate may be a little problematic to deal with, although I think both direct product could be isomorphic to a general uncountable direct product. Still, I can't see it that clear, yet. And "same number of $\;\Bbb Q$'s" is a little unaccurate here. Perhaps the problem is there, with "same cardinality" instead. – DonAntonio Nov 2 '13 at 20:15

Yes. A more simple counterexample is $\mathbb{Q}^{\oplus \mathbb{N}} \oplus \mathbb{Q} \cong \mathbb{Q}^{\oplus \mathbb{N}} \oplus 0$. Are you sure that the exercise in Hungerford is exactly as stated?
• Ok. I thought that I could reconstruct what Hungerford actually had in mind with this exercise, since I don't think that he would ask for a proof of a claim which is obviously false (after all, it's the same as in II.2.11 (c) but with $\mathbb{Q}$ instead of $\mathbb{Z}$!). But I'm not sure what Hungerford intended. Maybe the claim is true if $G,H,K$ are direct sums of finitely many "standard" divisible abelian groups ($\mathbb{Q}$ and $\mathbb{Z}/p^{\infty}$)? – Martin Brandenburg Nov 3 '13 at 9:33