Prove that a subgroup is cyclic Let $ H $ be the subgroup of the group $ (\mathbb R,+) $ such that $ H \cap [-1,1] $ is a finite set containing a nonzero element. Show that $H$ is cyclic.
 A: Hint: Write $H^+ = \{h \in H; h > 0\}$, assuming $H \neq 0$. And that if $m = \inf H^+ $, with $m \neq 0$ then $H = m \mathbb Z$.
By definition for every $\epsilon > 0$ there exists $h \in H^+$ such that $$m \leq h < m + \epsilon $$
Say $m \notin H$. Then you get a increasing sequence $(h_n)$ such that $$\lim h_n = m$$
as $H$ is a subgroup we have that $0 < h_n - h_{n-1} \in H^+$ thus $h_n - h_{n-1} > m$, which is a contradiction. Conclusion, $m \in H$ and it follows that $m\mathbb Z \subseteq H$. 
Conversely, let $h \in H^+$. Then there exists $q, r \in \mathbb Z$ such that $$h = qm + r$$
where $0 \leq r < m $. As $m= \inf H^+$ it must be that $r = 0$, because $r \in H$. Therefore $h = qm $ and then $H = m\mathbb Z$.
A: The finite set $H \cap [-1,1]$ has a nonzero element $h$ with minimal absolute value.
Try to prove that $h$ generates $H$.
Assume it does not. Pick $m \in H$ that is not in the subgroup generated by $h$.
By "approximating" $m$ by $h$, try to show that there is an element in $H$ with a smaller absolute value than $h$ contradicting the choice of $h$.
($m-nh \in H$ for all integers $n$)
This in turn proves that $H=(h)$.
I will give you more details for the approximation part.
I will do this through an example, as I think that you would get a lot
out of writing out the general case yourself.
Say $h=\frac{1}{2}$ is the element with minimal absolute value. We claim that $H=(\frac{1}{2})$. Assume not. Then, there is an element in $H$
that is not a multiple of $\frac{1}{2}$, say $\frac{13}{8}$. But, if
$\frac{1}{2}, \frac{13}{8}\in H$, then so is $\frac{13}{8}-\frac{1}{2}=\frac{9}{8} \in H$. Repeating the argument yields $\frac{9}{8}-\frac{1}{2}=\frac{5}{8} \in H$. Subtracting $\frac{1}{2}$ a last time yields
$\frac{1}{8} \in H$. This can not be true since $\frac{1}{8}$ has smaller absolute value than $\frac{1}{2}$. Hence, $\frac{13}{8}$ cannot be in $H$.
Now, try to genealize this proof. It might sill be tricky, but the argument is essentilly the same. Let me know how it goes.
