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.
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 $\,\displaystyle{x=\frac{a}{b}\in\Bbb Q}\,$ :
$$bp(x+H)=bp\left(\frac{a}{b}+H\right)=pa+H=(qa)\frac{p}{q}+H=qa\left(\frac{p}{q}+H\right)=qaH=H\Longrightarrow$$
$$\operatorname{ord}(x+H)\leq bp$$
And now finally (a) follows also from the above.
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.
