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

Question

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.

Answer 1

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.

Answer 2

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

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