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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 and here. This AMS reference 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 site. –  Zev Chonoles Dec 25 '12 at 15:45
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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? –  Zev Chonoles Dec 25 '12 at 15:47
    
Yeah, I'm sorry. I meant Q*. Thanks for correction. –  Michael Dec 25 '12 at 16:00

3 Answers 3

up vote 2 down vote accepted

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! –  Michael Dec 25 '12 at 16:07
    
I'm glad to be helpful. –  Ivan Dec 25 '12 at 19:32

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? –  Michael Dec 25 '12 at 16:08
    
yes, this will work with both groups. –  user29743 Dec 25 '12 at 16:10
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In the first group consider $\frac{1}{2^n}$. In the second, consider $2^{\frac{1}{n}}$. –  user29743 Dec 25 '12 at 16:10
    
Could you be clearer, countinghaus? –  Ivan Dec 25 '12 at 16:11
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A group which contains elements of arbitrarily large (finite) order cannot be cyclic. –  user26857 Dec 25 '12 at 18:08

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! +1 –  amWhy Mar 3 '13 at 0:11

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