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Suppose that $(\mathbb Q,+)$ be the additive rational group and $H$ a subgroup of it. Which of the following statements is not true?

(a) If $\mathbb Q/H\cong \mathbb Q$, then $H=0$.

(b) If $H\neq 0$, then every proper subgroup of $\mathbb Q/H$ is of finite order.

(c) If $H\neq 0$, then every element of $\mathbb Q/H$ is of finite order.

(d) If $\mathbb Q/H$ is finite, then $H=\mathbb Q$.

I have tried to solve it. I am not able to draw conclusions. Please help me.

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Add what you tried to the question. – rschwieb Jan 23 '13 at 15:02
Sorry. I'm a little confused – aliakbar Jan 23 '13 at 15:09
You mentioned that you tried to solve it: including your partial work is invaluable for us to help you. So... include your partial work! – rschwieb Jan 23 '13 at 15:18
@rschwieb What is the subgroups of $Q$? – aliakbar Jan 23 '13 at 15:20
(a) follows from (c). (c) and (d) use similar ideas. But again, post your own work! – user641 Jan 23 '13 at 15:50
up vote 3 down vote accepted

The additive group $\,\Bbb Q\,$ is as abelian divisible group, meaning:

$$\forall\,g\in \Bbb Q\,\,\wedge\,\,\forall n\in\Bbb N:=\{1,2,...\}\,\,\,\exists\,x\in \Bbb Q\,\,\,s.t.\,\,\,g=nx$$

Some facts that'd be interesting you can prove about divisible groups: if $\,G\,$ is a divisible group then

1) Every homomprphic image of $\,G\,$ is divisible

2) The only finite group that is divisible is the trivial group $\,\{0\}\,$

3) The group $\,G\,$ has no maximal subgroups

Thus, almost automatically we already have that (b) isn't true (and this already solves your problem), (d) is true.

Now, (c) follows from the following: if we take $\,\displaystyle{0\neq h=\frac{p}{q}\in H}\,$ then for any $\,\displaystyle{x=\frac{a}{b}\in\Bbb Q}\,$ :


$$\operatorname{ord}(x+H)\leq bp$$

And now finally (a) follows also from the above.

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+1 bbecause of ruling out the options nicely. – Babak S. Jan 24 '13 at 5:39

Hint: It can be easily verified that $\frac{\mathbb Q}{\mathbb Z}\cong\bigoplus_{p\in P}\mathbb Z(p^{\infty})$. So, I think we can work on $H=\mathbb Z\leq\mathbb Q$ and probe the right option.

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Helpful hint! +1 – amWhy Feb 7 '13 at 0:01

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