# $(\mathbb{Q},+)$ and $(\mathbb{Q} \times \mathbb{Q},+)$ are not isomorphic as groups

I need help in solving this assignment question in abstract algebra.

Prove that $$(\mathbb{Q},+)$$ and $$(\mathbb{Q} \times\mathbb{Q},+)$$ are not isomorphic as groups .

I am unable to find a property that one of group would satisfy but not other despite thinking a lot .

Kindly help.

Thanks!!

• Compare also with this post, and this one. Aug 15, 2020 at 16:19

In $$(\mathbb Q,+)$$, every finitely-generated subgroup is cyclic. In $$(\mathbb Q\times\mathbb Q,+)$$, the subgroup $$(\mathbb Z\times\mathbb Z, +)$$ is finitely generated but not cyclic - it is of rank $$2$$.

– YCor
Aug 16, 2020 at 17:57
• It is implicitly by contradiction. Mar 21, 2022 at 14:07

Suppose there is an isomorphism $$f:\mathbb Q\to\mathbb Q\times\mathbb Q$$. Let $$x\in\mathbb Q$$ be such that $$f(x)=(0,1)$$ and $$y\in\mathbb Q$$ be such that $$f(y)=(1,0).$$

Since $$x$$ and $$y$$ must be nonzero rational numbers, there are integers $$n,m\in\mathbb Z$$ which are not both zero such that $$nx+my=0$$. However, this would mean $$(n,m)=nf(x)+mf(y)=(0,0)$$, which is a contradiction. Thus, no such $$f$$ exists.

Let $$\phi:\mathbb{Q}\longrightarrow\mathbb{Q}\times\mathbb{Q}$$ be an isomorphism and let $$H=\phi(\mathbb{Z})$$. Then $$\phi$$ should induce an isomorphism $$\bar\phi:\frac{\mathbb{Q}}{\mathbb{Z}}\longrightarrow\frac{\mathbb{Q}\times\mathbb{Q}}{H}.$$ But $$\frac{\mathbb{Q}}{\mathbb{Z}}$$ is torsion while $$\frac{\mathbb{Q}\times\mathbb{Q}}{H}$$ is not as easily checked.

Assume they were isomorphic. Let $$\varphi:\mathbb{Q}\to\mathbb{Q}^2$$ be such an isomorphism. Hence $$\varphi(1)=(x,y)$$ for some $$x,y\in \mathbb{Q}$$. Then $$\varphi(r)=(rx,ry)$$ for all $$r\in\mathbb{Q}$$ (do you understand why?). We know that $$x\neq 0$$ and $$y\neq 0$$ (otherwise $$(1,1)$$ wouldn't be in the image of $$\varphi$$).

But now we see that $$(2x,y)$$ cannot be in the image of $$\varphi$$: If $$\varphi(r)=(2x,y)$$, then $$(rx,ry)=(2x,y)$$, hence $$2=r=1$$, which is a contradiction.

We conclude that $$\varphi$$ cannot exist.