$\mathbb{R}/H$ has a nontrivial element of finite order I am currently struggling with this problem:

Let $H$ be a nontrivial subgroup of $\mathbb{R}$. Prove that $\mathbb{R}/H$ has a nontrivial element of finite order.

I believe this refers to the additive group of real numbers, given that it is represented with $\mathbb{R}$ rather than $\mathbb{R} \backslash \{0\}$ or $\mathbb{R}^*$. Thus, if we take 
$$
\mathbb{R}/H:=\{\,r+H \mid r \in \mathbb{R}\,\}
$$ 
it seems that the problem boils down to finding some $r+h \in R/H$ and some $n \in \mathbb{z}$ for which $(r+h)+...+(r+h)_n=0$. I've toyed around with the algebra, but I just can't seem to come up with a proof. Any help would be greatly appreciated!
 A: Let $H=\Bbb Q$. Then an element $a+\Bbb Q$ of finite order $n$ in $\Bbb R/\Bbb Q$ means that $na\in \Bbb Q$. But then $a\in\Bbb Q$ and $a+\Bbb Q$ is trivial. So the claim is false.
A: This is not quite correct, you need to find an integer $n$ and an $r\in\mathbb{R}$ that is not in $H$ such that $nr\in H$. Also I'm not sure where you got 1 from exactly since the identity in the additive group is 0. 
Now suppose we have an $H$ which doesn't have this property, then if $h\in H$, and $q\in \mathbb{Q}$, if $qh \not\in H$, then take the denominator of $q$, $d$, and $d(qh)=sh\in H$ where $s$ is the numerator of $q$. Thus $H$ must be closed under scaling by rationals. Thus $H$ is a $\mathbb{Q}$ vector space.
But take a $\mathbb{Q}$-vector subspace of $\mathbb{R}$, $H$. By definition these are abelian groups, and in addition they are closed under scaling by rationals. Now suppose $\mathbb{R}/H$ had an element of finite order, then $nr \in H$ for some $n\in\mathbb{Z}$, but since $H$ is a $\mathbb{Q}$ vector space, we have $\frac{1}{n}\cdot(nr)=r\in H$, so this element of finite order is trivial.
Thus the subgroups $H$ of $\mathbb{R}$ such that $\mathbb{R}/H$ has no element of finite order are precisely the $\mathbb{Q}$-vector subspaces of $\mathbb{R}$, and as Hagen von Eitzen pointed out, an example of such a space is $\mathbb{Q}$, so the claim is false.
