Cyclicity of some special subgroup of $(\mathbb R,+)$

Question

If $H$ is a subgroup of $(\mathbb R,+)$ such that $H\cap[-1,1]$ is finite and contains a positive element then is it true that $H$ is cyclic ?

You could try to prove the basic: Any subgroup of $;\Bbb R;$ is either cyclic or dense in $;\Bbb R;$ . This answers your question at once. — Timbuc, Nov 15 '14 at 13:20
Does this answer your question? Discrete subgroups of $(\mathbb{R},+)$ or this https://math.stackexchange.com/questions/90177/subgroup-of-mathbbr-either-dense-or-has-a-least-positive-element — Anne Bauval, Nov 07 '22 at 06:52

score 4 · Accepted Answer · edited Jan 17 '23 at 10:58

4

Given that $H\cap [-1,1]$ is finite. Then, the set contains a least positive element. Let's say $a$ is this element.

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

Suppose, $\exists b\in H$ such that $b$ not in $\langle a\rangle$.

Apply division algorithm and try to get to a contradiction.

edited Jan 17 '23 at 10:58

Anne Bauval

answered Nov 15 '14 at 12:56

Swapnil Tripathi

1

I think this proof doesn't work. Consider $H=2\Bbb{Z}$. In that case, we get $H\cap[-1,1]={0}$. – Hanul Jeon Nov 15 '14 at 12:59
Sorry, I misunderstand the question. – Hanul Jeon Nov 15 '14 at 13:05
@SwapnilTripathi: If I am not misunderstanding , neither $b$ nor $a$ should be integers , how do I apply division algorithm ? – Nov 15 '14 at 13:08
Actually, I shouldn't say division algorithm. But note that if there exists such a $b\in H\bigcap[-1,1]$ as in the answer, then of course for some $n\in\mathbb{N}$, $na<b<(n+1)a$, thus $0<b-na<a$ and this contradicts the minimality of $a$. This looks somewhat like the division algorithm. – Swapnil Tripathi Nov 15 '14 at 13:32
1

I think $a$ should be chosen as 'the least positive element', instead of the 'minimum element'; and your claim should read $H = $ – user49685 Nov 15 '14 at 13:57
Yes, my bad! Edited accordingly. :) – Swapnil Tripathi Nov 15 '14 at 14:04
1

@AnneBauval Hi. I opened the site after a long time, apologies for the delay. I just realised that the first comment made by Hanul Jeon does not satisfy the hypothesis of the question since the intersection does not have a positive element. – Swapnil Tripathi Jan 17 '23 at 10:35

Jihad · Answer 2 · 2014-11-15T13:10:43.647

-1

Suppose that $H$ is not cyclic and it has at least $2$ $\mathbb{Q}$-independent generators $g_1, g_2$.
Without loss of generality one of them $g_1 = 1$ (otherwise we can assume subgroup $g_1^{-1}H$ and it isomorphic to $H$). From this we can say that $g_2$ is irrational.
Finally, $\forall x \in \mathbb{R} \backslash \mathbb{Q}$ and $ \forall m,n \in \mathbb{N}$ $nx \neq mx$ (mod $1$) (otherwise $x$ is rational). So all residues $nx$ modulo $1$ are different. Contradiction with fact that $H \cap [-1, 1]$ is finite.

edited Nov 15 '14 at 13:10

answered Nov 15 '14 at 13:05

Jihad

Point 1 is very incomplete (why does non-cyclicity imply the existence of two such elements?) and the last sentence of point 3 is mysterious. Moreover, the method for the wlog of point 2 is incompatible with the hypothesis $H \cap [-1, 1].$ – Anne Bauval Nov 07 '22 at 06:37

2 Answers2