# Can $\mathbb{Q×Q}$ be embedded in $\mathbb{R}$ as group?

I think ans is NO : if possible let that is true hence there is a monomorphism from $$H= \mathbb{Q×Q}$$ to $$\mathbb{R}$$. as $$\mathbb{R}$$ has only subgroups which is cyclic or dense and $$H$$ is not cyclic hence dense but it's proper subgroup $$\mathbb{Z×Z}$$ is neither cyclic nor dense in $$\mathbb{R}$$ hence contradiction. Hence the claim.

Is my proof correct??
Thanks.

• How do you know $\mathbb{Z}\times\mathbb{Z}$ is not dense in $\mathbb{R}$? Jul 6, 2019 at 3:54
• Try $(a,b)\mapsto a+b\sqrt2$. Jul 6, 2019 at 3:56
• Your proof is incorrect. While you claim to have derived a contradiction, you did not actually arrive at any contradiction. Note that while $\Bbb Z\times\Bbb Z$ may not be dense within $\Bbb Q\times\Bbb Q$ with the product topology, that doesn't mean its isomorphic copy in $\Bbb R$ isn't dense. Jul 6, 2019 at 4:22
• If you know that $\mathbb{R}$ is an infinite-dimensional vector space over $\mathbb{Q}$ then the answer is trivially yes. Jul 6, 2019 at 4:35
• The question has nothing to do with topology—only the algebraic structures of the groups—so arguments about density are unlikely to be relevant. Jul 6, 2019 at 4:44

Let $$\{e_i: i \in I\}$$ be a basis for the vector space of $$\Bbb R$$ over the field $$\Bbb Q$$. It is clear from cardinality considerations that $$I$$ is uncountable.

Pick any two distinct elements from the base, say $$e_{i_1}$$ and $$e_{i_2}$$. Map $$(q,q') \in \Bbb Q^2$$ to $$qe_{i_1} + q'e_{i_2} \in \Bbb R$$ and note that we have group embedding.

This of course works for any finite power of the rationals.

The map $$(a,b)\mapsto a+b\sqrt2$$ is an injection $$\mathbb{Q×Q} \to \mathbb{R}$$ because $$\sqrt2$$ is irrational.

$$\sqrt2$$ is not special here; any irrational number works, for instance $$\pi$$.

• Based on a comment by Lord Shark the Unknown.
– lhf
Dec 23, 2019 at 18:50