\mathbb{Q}\times\mathbb{Q} is not cyclic

**dwsmith** · October 8th 2011, 06:45 PM

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

How do I do this my book says nothing.

**Drexel28** · October 8th 2011, 06:49 PM

Originally Posted by dwsmith

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

How do I do this my book says nothing.

If $\mathbb{Q}^2$ was cyclic, then $\mathbb{Q}$ being the image of $\mathbb{Q}^2$ under the projection mapping would be cyclic, so it suffices to show that $\mathbb{Q}$ is not cyclic. To see this suppose that there was an isomorphism $f:\mathbb{Q}\to\mathbb{Z}$ we see then $f(1)=f(m\frac{1}{m})=mf(\frac{1}{m})$ so that $f(1)$ is divisible by $m$ for every $m\in\mathbb{Z}$ and so $f(1)=0$ , but since $f(0)=0$ this contradicts injectivity of $f$ .

**Swlabr** · October 10th 2011, 01:25 AM

Originally Posted by Drexel28

If $\mathbb{Q}^2$ was cyclic, then $\mathbb{Q}$ being the image of $\mathbb{Q}^2$ under the projection mapping would be cyclic, so it suffices to show that $\mathbb{Q}$ is not cyclic.

Can;t you just say "subgroups of cyclic groups are cyclic"...

**roninpro** · October 10th 2011, 11:48 AM

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

**Drexel28** · October 10th 2011, 01:22 PM

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

Another good way to look at it! My method shows more generally that any divisible group is not cyclic.

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