2
$\begingroup$

а) $\mathbb{Q}/2\mathbb{Z}$

b) $\mathbb{R}^*/\mathbb{Q}^*$

How to find out (and prove) if the factor groups above are cyclic or not?

Thanks in advance!

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    I don't know if I interpreted your question correctly in making my edits, so I ask you to check that I didn't make any mistakes. You can find some good starting points on how to format mathematics on the site [here](http://meta.math.stackexchange.com/a/464/264) and [here](http://meta.stackexchange.com/a/70559/161783). [This AMS reference](ftp://ftp.ams.org/ams/doc/amsmath/short-math-guide.pdf) is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own [TeX.SE](http://tex.stackexchange.com/) site.2012-12-25
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    If I did interpret your question correctly, part b doesn't make sense since $\mathbb{Q}$ is not a subset of $\mathbb{R}^*=\mathbb{R}\setminus\{0\}$. Do you mean $\mathbb{Q}^*$ perhaps?2012-12-25
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    Yeah, I'm sorry. I meant Q*. Thanks for correction.2012-12-25

3 Answers 3

2

Suppose that $\mathbb{Q}/ 2 \mathbb{Z}$ is cyclic $\Rightarrow \exists a=m/s \in \mathbb{Q}: a^k + 2 \mathbb{Z} = 1/2 + 2 \mathbb{Z} \Rightarrow ka - 1/2 \in 2 \mathbb{Z} \Rightarrow \exists n \in \mathbb{Z} : ka-1/2=2n \Rightarrow 2ka-1=4n \Rightarrow (2km)/s -1 = 4n \Rightarrow 2km -s = 4ns$ which seems to me to be clearly impossible . [I think I'll have to edit this demonstration]

For the second problem consider $I := <\pi,e>$. $\pi_{\mathbb{Q}}(I)=<\pi,e> \mathbb{Q} .$ $ \pi_{\mathbb{Q}}(I) < \mathbb{R}^*/\mathbb{Q}^*$ because it's the imagine of a subgroup. If $\mathbb{R}^*/\mathbb{Q}^*$ is cyclic $\Rightarrow \pi_{\mathbb{Q}}(I)$ is cyclic but if it's cyclic it's generate by a $x\mathbb{Q}$ for such $x \in \mathbb{R}^*$. So $x^n\mathbb{Q} = e\mathbb{Q} \Rightarrow x^n/e \in Q \Rightarrow e|x.$ Same idea proves that $\pi|x$ So $ x=\pi e r$ (with $r \in \mathbb{R}$) but it's clear that $ \forall n,$ $ (\pi e r)^n\mathbb{Q}\not=e \mathbb{Q} $ , becuase $\pi \not\in \mathbb{Q}$.

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    Thanks a lot, you truly helped me!2012-12-25
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    I'm glad to be helpful.2012-12-25
1

I think the first group is not cyclic, because it is divisible, torsion group and so:

$$\frac{\mathbb Q}{2\mathbb Z}\cong \sum_p\mathbb Z(p^{\infty})$$

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    Good observations! +12013-03-03
1

Hint: a group which contains torsion elements of arbitrarily large order cannot be cyclic. (Proof: in order to get a contradiction, suppose $t$ is a generator. Show that $t$ must be torsion. What is its order?)

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    Thanks, I think I got the idea. Are you talking about both groups?2012-12-25
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    yes, this will work with both groups.2012-12-25
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    In the first group consider $\frac{1}{2^n}$. In the second, consider $2^{\frac{1}{n}}$.2012-12-25
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    Could you be clearer, countinghaus?2012-12-25
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    A group which contains elements of arbitrarily large (finite) order cannot be cyclic.2012-12-25