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$\mathbf{Question:}$ $(H,+)$ 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<h-pa_t<a_t $ which contradicts the fact that $a_t$ is the minimal element in $Q$.

Is this Proof valid? Kindly verify.

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    $\begingroup$ Your proof looks good to me. $\endgroup$ – Rob Arthan Jun 23 at 22:00
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Looks good to me. Maybe add an argument why $Q$ is not empty (I am sure that you understand why it is non-empty).

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  • $\begingroup$ $T$ being non-empty, there is some $-1 \leq z<0 $. Therefore, $0<-z \leq 1$, which is in $Q$.Is it correct? $\endgroup$ – Subhasis Biswas Jun 23 at 22:05
  • $\begingroup$ 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 $-z \in Q$. $\endgroup$ – ThorWittich Jun 23 at 22:10

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