If $(H,+)$ is a subgroup of $(\mathbb{R},+)$ with finite $H \cap [-1,1]$ with some non-zero element, then $H$ is cyclic.

$$\mathbf{Question:}$$ $$(H,+)$$ is a subgroup of $$(\mathbb{R},+)$$ such that $$H \cap [-1,1]$$ is finite and contains elements other than $$0$$. Show that $$(H,+)$$ must be cyclic.

$$\mathbf{Attempt:}$$ Since $$\{0\}\cup T= H \cap [-1,1]$$ is finite [$$0 \not\in T$$, $$\emptyset \subsetneq T$$], so we can completely enumerate $$T$$. Let $$T =\{a_1, a_2, ..., a_m\}$$ be the complete 'list' of elements. Let $$Q= \{a_i \in T: a_i>0 \}$$. Let $$a_t$$ be the minimal element in $$Q$$.

Claim: $$H=\langle a_t \rangle$$.

Suppose, the claim is false. Then for some $$h \in H$$, $$h \notin \langle a_t \rangle$$. [we pick $$h$$ such that it is $$>0$$. If not, then we pick $$-h$$].

So, we can find a positive integer $$p$$ such that $$p a_t< h <(p+1)a_t \implies 0 which contradicts the fact that $$a_t$$ is the minimal element in $$Q$$.

Is this Proof valid? Kindly verify.

• Your proof looks good to me. – Rob Arthan Jun 23 at 22:00

Looks good to me. Maybe add an argument why $$Q$$ is not empty (I am sure that you understand why it is non-empty).
• $T$ being non-empty, there is some $-1 \leq z<0$. Therefore, $0<-z \leq 1$, which is in $Q$.Is it correct? – Subhasis Biswas Jun 23 at 22:05
• That is the non-trivial case, yes. There is one element $z \in T$. Thus by definition $z > 0$ or $z < 0$. If $z > 0$, then we have $z \in Q$ and if $z < 0$ we have $-z \in T$ with $-z > 0$, such that $-z \in Q$. – ThorWittich Jun 23 at 22:10