Let $H$ be a subgroup of the group $(R, +)$ such that $H$ $∩$ [-1,1] is a finite set containing a non zero element. Show that $H$ is cyclic. Observations:
Since $H$ is a subgroup of $(R, +)$ so $0 \in H.$ 
If $1 \in H,$ then all positive integers belong to $H.$ But $H$ is closed wrt addition, so the negative integers must belong to $H$ as well. Similarly, for $-1.$ Thus if $1$ or $-1 \in H$, then the whole set of integers belongs to $H.$ 
If $1/n$ is in $H$ (where $n$ is a positive integer), then $n.(1/n) = 1$ also belongs to $H,$ which in turn ensures that all integers belong to $H.$
Something similar can be said about $-1/n$.
So it is observed that $H$ will contain the entire set of integers.
But how do I conclude that in every case $H$ is isomorphic to $\mathbb Z$, and hence is cyclic?
 A: Hints:
Let $\;h\;$ be minimal wrt $\;\begin{cases}h\in H\\0<h<1\end{cases}\;\;\;$ , and let
$\;x\in H\;$  .
We can write $\;x=mh+r\;,\;\;m\in\Bbb Z\;,\;\;0\le r<h\;$ . Deduce it must be $\;r=0\;$ otherwise you get a contradiction to minimality.
A: Let's take the setting in Timbuc's answer (which, by the way, provides a very nice hint), but removing the requisite that $h<1$. So $h$ is the least positive element of $H$ and we know that this exists since $\min [(H \cap (0,1])\cup \{1\}]$ is a lower bound.
So take $x\in H$. Let $r$ be least nonnegative  element in the set 
$$
R:= \{x - mh : m\in \mathbb{Z}\}
$$
We claim that $r=0$, and then we would have $H=\langle h \rangle_{\mathbb{Z}}$.
Since for some $m\in\mathbb{Z}$, $r = x - mh$, we have $r\in H$. Moreover, $r<h$ since otherwise $0\leq r-h=x-(m+1)h$ would be a smaller nonnegative element of $R$. Hence by minimality of $h$ we must have $r=0$.
A: Hint!
Think about the intersection. It is finite so if you "take out" the $0$, you can now take a minimal element, let's call it $x$.
Now if all the other elements aren't a multiple of $x$, think why we would have a contradiction with it's minimality.
