If $G \oplus H$ is isomorphic to a proper subgroup of itself, then must the same be true of one of $G$ and $H$? 
Let $G$ and $H$ are groups. If $G \oplus H$ is isomorphic to a proper subgroup of itself, then must the same be true of one of $G$ and $H$?

I found some examples of $G$ such that $G$ has no proper subgroup isomorphic to $G$.
For example, $\mathbb{Q}$ and $\mathbb{Q}\oplus \mathbb{Q}$ has no proper  subgroup isomorphic to each mother group.
(The reason: If $f:\mathbb{Q}\oplus \mathbb{Q}\rightarrow\mathbb{Q}\oplus \mathbb{Q}$ is injective group homomorphism, then $f$ is also $\mathbb{Q}$-module homomorphism, so $f$ is $\mathbb{Q}$-linear map. So, injectivity of $f$ implies surjectivity of $f$. This means $\mathbb{Q}\oplus \mathbb{Q}$ has no isomorphic subgroup.)
I think there is counter-example of this claim but I can't choose one..
How to prove or take counter-example?
 A: This seems to be an open question. Lets start by rewriting it. A group $G$ is called coHopfian if any injective endomorphism $G\to G$ is an isomorphism. Then the question asks:

Question. Suppose $A$ and $B$ are coHopfian. Is $A\times B$ necessarily coHopfian?

(Thanks to Jeremy Rickard for correcting my interpretation of this problem in the comments.)
The paper Y. Li, On the cohopficity of the direct product of cohopfian groups. Comm. Algebra 35 (2007), no. 10, 3226–3235 (doi), addresses this question. They write "The question was raised by Yongkuk Kim, and informed to the author by Professor Shicheng Wang", and they prove the following.
A group $G$ is extremely noncommutative if $[g, h]=1$ implies $\langle g, h\rangle$ is cyclic. For example, torsion-free hyperbolic groups are extremely noncommutative (so then "almost all" finitely presented groups satisfy this property).
Theorem 1.
Suppose both $A$ and $B$ are coHopfian groups. Then $A\times B$ is coHopfian if $A$ is either:

*

*Extremely noncommutative and torsion free; or

*Finite Abelian; or

*Finite simple.

They also prove the following interesting result.
Proposition 1.
Let $A$ and $B$ be two coHopfian groups. If there does not exist nontrivial homomorphism from $A$ to $B$ or from $B$ to $A$, then $A\times B$ is coHopfian.
The paper B. Goldsmith and K. Gong, Algebraic entropies, Hopficity and co-Hopficity of direct sums of Abelian groups. Topol. Algebra Appl. 3 (2015), no. 1, 75–85 (doi), also addresses this question. They improve on parts 2 & 3 of Theorem 1:
Theorem 4.10.
Suppose that $A$ is a coHopfian group and $B$ is a finite group, then $A\times B$ is coHopfian.
They also prove the following, which improves on part 2 of Theorem 1 in a different direction:
Theorem 4.11.
Suppose that $A$ is a coHopfian group and $B$ is a finitely co-generated Abelian group. Then $A\times B$ is coHopfian.
