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

Question

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!!

Answer 1

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 $2$.

Answer 2

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.

Answer 3

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.

Answer 4

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.