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My question is if

Is it possible to have $G\oplus H \simeq G$ for groups $G$ and $H$ abelian groups?

I know this is possible for $H$ the trivial group and I can see that if $G$ and $H$ are finite, then there are no other examples because of the orders of the groups.

From Does $G\oplus G$=$H\oplus H$ imply $G$=$H$ for divisible abelian groups? it sounds like one would need a non divisible group.

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John Doe
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  • There is a divisible example, as I pointed out on T. Bongers' answer. – Dustan Levenstein Aug 02 '16 at 20:41
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    It is worth pointing out that by the structure theorem for finitely generated abelian groups (corollary of the structure theorem for finitely generated modules over PIDs), there are no finitely generated examples. Therefore, you aren't likely to find anything substantially simpler than the answer provited by T. Bongers. – Aaron Aug 02 '16 at 21:22
  • @Aaron: If you would elaborate on this comment in an answer, then I will accept your answer. – John Doe Aug 03 '16 at 12:12

2 Answers2

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$$G = H = \prod_{n \in \mathbb{N}} \mathbb{Z}$$

or $G$ as above, $H = \mathbb{Z}$.

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    +1. Note that by using $\bigoplus\mathbb{Z}$ instead, we can get a countable example. – Noah Schweber Aug 02 '16 at 20:36
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    Note that replacing $\mathbb Z$ with $\mathbb Q$ gives a divisible example. OP's linked question is not relevant - that question is about when $G \oplus G \simeq H \oplus H$, not when $G \oplus H \simeq G$. – Dustan Levenstein Aug 02 '16 at 20:40
  • It is vaguely relevant in the sense that if you have a group $G$ such that $G\oplus G\not\cong G$ but $G\oplus G\oplus G\cong G$ then you can answer both questions in the negative with $H=G\oplus G$. See the comments to this answer – Myself Aug 02 '16 at 23:35
  • @NoahSchweber: Is $\prod$ the same as $\oplus$ here? – John Doe Aug 03 '16 at 12:13
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    @JohnDoe It isn't. While finite products are the same as finite coporoducts of abelian groups, you run into differences when you are dealing with an infinite indexing set. Without getting into generalities, $\bigoplus_{\mathbb N} \mathbb Z$ is the collection of sequences of integers which are eventually zero, while $\prod_{\mathbb N} \mathbb Z$ is the collection of all sequences of integers. – Aaron Aug 03 '16 at 12:42
  • @Aaron: Ahh. ok and the point is that you can use both to get an example? – John Doe Aug 03 '16 at 12:44
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    @JohnDoe Yes, you can use either to get an example, but with $\bigoplus$ the example is countable, which is in some sense nicer (in that it shows you don't have to deal with huge objects, just infinite ones). – Aaron Aug 03 '16 at 12:45
  • @Aaron: Ok, thanks for this. If you write this up in a nice answer, I will accept it. – John Doe Aug 03 '16 at 12:46
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If $G$ is a finite group, then, as you noted, simply looking at the cardinality of $G$ as a set shows there are no examples except when $H$ is trivial. By a similar token, if we have an infinite set $X$, then $X$ will be in bijection with a proper subset of $X$, and we can upgrade this to an example, e.g., $\bigoplus_{\mathbb N} \mathbb Z \cong \mathbb Z \oplus \bigoplus_{\mathbb N} \mathbb Z$.

This leaves open the question is whether we can find "smaller" examples, where we aren't finite as sets, but are still finitely generated as groups. The answer is no. As I suggested in my comment, one can show this using the structure theorem for finitely generated abelian groups. However, I would like to present a slightly different proof that doesn't use the full power of the classification. However, if you follow it with the classification in mind, you will see that it is just breaking the group into it's finite and infinite components, and comparing those separately.

Given an abelian group $G$, define the torsion subgroup $T(G):=\{g\in G \mid \exists n\in \mathbb N \text{ such that } ng=0\}$. I leave it as an exercise to show that

  • $T(G\oplus H)\cong T(G)\oplus T(H)$.
  • If $G=T(G)$, and $G$ is finitely generated, then $G$ is finite.

So suppose that $G \cong G\oplus H$, with $G$ finitely generated. Then examining torsion subgroups, $T(G)\cong T(G)\oplus T(H)$, and since $T(G)$ is finite, we must have $T(H)=\{0\}$.

Given an abelian group $G$, the tensor product $V(G):=G\otimes_{\mathbb Z}\mathbb Z$ is a $\mathbb Q$-vector space. Here, $V(G)$ is nonstandard notation, to be read "the vector space of $G$". If $\{g_1, \ldots, g_n\}$ are generators for $G$, then the elements $g_i\otimes 1$ span $V(G)$, and so if $G$ is finitely generated, then $V(G)$ is finite dimensional. Because tensor products commute with direct sums, $V(G\oplus H)\cong V(G)\oplus V(H)$. Therefore, if $G\cong G\oplus H$, then $V(G)\cong V(G)\oplus V(H)$, and since the dimension of a vector space is well defined and additive, $V(H)\cong \{0\}$.

Putting everything together, if $G$ is finitely generated and $G\cong G\oplus H$, then $T(H)$ is the trivial group and $V(H)$ is trivial vector space. This implies (exercise!) that $H$ is the trivial group.

Aaron
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  • The last "(exercise!)" doesn't look very easy to me if you don't have access to the classification of finitely generated abelian groups... – Najib Idrissi Aug 03 '16 at 13:49
  • @NajibIdrissi Maybe I'm thinking about things too casually, but you just need to show that if $g\in G$ is non-torsion, then $g\otimes 1$ is nonzero. If this is too messy, then also follows from flatness of $\mathbb Q$ as a $\mathbb Z$-module, although this is probably not a useful perspective for an undergraduate to take. But the statement doesn't require $H$ to be finitely generated, and I was only offering this as an alternative to using the classification theorem, which I think is the simplest route. – Aaron Aug 03 '16 at 14:01
  • Well that's the thing, how do you show that $g \otimes 1$ is nonzero? – Najib Idrissi Aug 03 '16 at 14:05
  • @NajibIdrissi If you are in a commutative algebra class, you recognize the tensor product as a localization, and then in $S^{-1}M$, $x/1\sim 0/1 \Leftrightarrow \exists s\in S \text{ s.t. } sx=0$ and so $g\otimes 1=0$ if and only if $g$ is torsion, but how you approach it otherwise will depend on what you've seen. – Aaron Aug 03 '16 at 14:33