# Thread: \mathbb{Q}\times\mathbb{Q} is not cyclic

1. ## \mathbb{Q}\times\mathbb{Q} is not cyclic

Prove that $\displaystyle \mathbb{Q}\times\mathbb{Q}$ is not cyclic.

How do I do this my book says nothing.

2. ## Re: \mathbb{Q}\times\mathbb{Q} is not cyclic

Originally Posted by dwsmith
Prove that $\displaystyle \mathbb{Q}\times\mathbb{Q}$ is not cyclic.

How do I do this my book says nothing.
If $\displaystyle \mathbb{Q}^2$ was cyclic, then $\displaystyle \mathbb{Q}$ being the image of $\displaystyle \mathbb{Q}^2$ under the projection mapping would be cyclic, so it suffices to show that $\displaystyle \mathbb{Q}$ is not cyclic. To see this suppose that there was an isomorphism $\displaystyle f:\mathbb{Q}\to\mathbb{Z}$ we see then $\displaystyle f(1)=f(m\frac{1}{m})=mf(\frac{1}{m})$ so that $\displaystyle f(1)$ is divisible by $\displaystyle m$ for every $\displaystyle m\in\mathbb{Z}$ and so $\displaystyle f(1)=0$, but since $\displaystyle f(0)=0$ this contradicts injectivity of $\displaystyle f$.

3. ## Re: \mathbb{Q}\times\mathbb{Q} is not cyclic

Originally Posted by Drexel28
If $\displaystyle \mathbb{Q}^2$ was cyclic, then $\displaystyle \mathbb{Q}$ being the image of $\displaystyle \mathbb{Q}^2$ under the projection mapping would be cyclic, so it suffices to show that $\displaystyle \mathbb{Q}$ is not cyclic.
Can;t you just say "subgroups of cyclic groups are cyclic"...

4. ## Re: \mathbb{Q}\times\mathbb{Q} is not cyclic

To show that $\displaystyle \mathbb{Q}$ is not cyclic, it seems conceptually more straightforward to take an arbitrary number $\displaystyle m/n$ (say, in lowest terms) and say that it can never meet $\displaystyle 1/p$, where $\displaystyle p$ is some prime that is not a factor of $\displaystyle n$. (For example, no multiples of $\displaystyle 3/16$ will ever be $\displaystyle 1/5$.) So no single number can generate all of $\displaystyle \mathbb{Q}$.

5. ## Re: \mathbb{Q}\times\mathbb{Q} is not cyclic

Originally Posted by Swlabr
Can;t you just say "subgroups of cyclic groups are cyclic"...
Sure, either way works.

Originally Posted by roninpro
To show that $\displaystyle \mathbb{Q}$ is not cyclic, it seems conceptually more straightforward to take an arbitrary number $\displaystyle m/n$ (say, in lowest terms) and say that it can never meet $\displaystyle 1/p$, where $\displaystyle p$ is some prime that is not a factor of $\displaystyle n$. (For example, no multiples of $\displaystyle 3/16$ will ever be $\displaystyle 1/5$.) So no single number can generate all of $\displaystyle \mathbb{Q}$.
Another good way to look at it! My method shows more generally that any divisible group is not cyclic.