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Prove that the product of two infinite cyclic groups is not infinite cyclic.

Let $G=<a>$ and $G'=<b>$. The product group is $\{a^mb^n: m,n\in\mathbb{Z}\}=<a^kb^l>$

Now, for given $m,n\in\mathbb{N}$, we can find $u$ such that $(a^kb^l)^u=a^mb^n$. Then $u=\frac{m}{k}=\frac{n}{l}\Rightarrow \frac{k}{l}=\frac{m}{n}$, which is true for any $m,n\in\mathbb{N}$, which can not be true, since $k,l$ are fixed, so the ratio is. And then I am stuck. What to do next???

Another question from my curiosity: We can take $G=<e^{i\pi \sqrt{2}}>$ and $G'=<e^{i\pi \sqrt{3}}>$. What is the product group? It should be $<e^{i\pi \theta}>$, where $\theta$ is rational. But what is $\theta$?

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    $\begingroup$ Do you mean the product group to be $\langle a, b \mid ab=ba \rangle$? That is, the direct product of the two cyclic groups $\langle a \rangle$ and $\langle b \rangle$. $\endgroup$
    – Randall
    Aug 1, 2017 at 16:41
  • $\begingroup$ Are $G$ and $G'$ subgroups of the same group (as the notation inside $\{\cdots\}$ would suggest)? If not, what do you mean by "product" and how does it fare with the (non-standard for a non-embedded case, in my opinion) notation $a^mb^n$? Also, perhaps you mean $m,n\in \Bbb Z$. $\endgroup$
    – user228113
    Aug 1, 2017 at 16:42
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    $\begingroup$ At any rate, it isn't too hard to show that $\mathbb{Z} \times \mathbb{Z}$ isn't cylic by picking an arbitrary element, examining what it generates, and seeing that it must miss something. You may need to argue on cases on this generating element $(a,b)$ by considering (1) $a=0$, (2) $b=0$, and (3) neither equal to 0. $\endgroup$
    – Randall
    Aug 1, 2017 at 16:45
  • $\begingroup$ Your so called product group isn't very clear. Can you be a little more precise? $\endgroup$
    – Naive
    Aug 1, 2017 at 16:46

1 Answer 1

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The term product in the question probably means cartesian product. Since $\mathbb{Z}$ is the only infinite cyclic group up to isomorphism, the question reduces to:

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

Here is a roadmap:

  • The image of a cyclic group under a homomorphism is cyclic

  • $C_2 \times C_2$ is a homomorphic image of $\mathbb{Z} \times \mathbb{Z}$

  • $C_2 \times C_2$ is not cyclic

Here, $C_2$ is the cyclic group of order $2$.

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  • $\begingroup$ Thanks for the answer!!! $\endgroup$
    – MAN-MADE
    Aug 1, 2017 at 18:14

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