Let-H-be-a-subgroup-of-R-such-that-H-1-1-contains-a-non-zero-element-Prove-that-H-is-cyclic-

Question Number 193794 by Rajpurohith last updated on 20/Jun/23

L e t H b e a s u b g r o u p o f (R, +) s u c h t h a t H \cap [- 1, 1]

c o n t a i n s a n o n z e r o e l e m e n t .

P r o v e t h a t H i s c y c l i c .

Answered by TheHoneyCat last updated on 28/Jun/23

Sorry but I think you forgot some extra hypothesis... In its current form, your problem can′t be solved. It is very easy to see so by considering: H=(R,+) Its intersection with [−1,1] DOES contain a non −zero element (say (π/4) or (1/2^n ) with n your favorite integrer) however R is not cyclic. To proove it consider x_0 the element that would generate it. Then ∀x∈R ∃n∈Z∣ x=nx_0 But ∄n∈Z∣ nx_0 =(x_0 /2)∈R There are several possible alternative questions you might have wanted to ask: 1) H∩[−1,1] contains no non−zero element 2) H∩[−1,1] contains finitely many elements Here are quick proofs of the two alternatives let x_0 =Inf H∩R_+ ^∗ Wether in case 1) or 2) if x_0 is not in H, then by definition of Inf, we can construct a sequence y_n ∈H ∣ y_(n ) →_(n→∞) x_0 and y_n −y_(n+1) →0 Meaning H cannot be cyclic (supose it is generated by z_0 then ∀x∈H ∃k∈Z∣ kz_0 =x but the taking n such that 0<∣y_n −y_(n+1) ∣<∣z_0 ∣ y_n −y_(n+1) ∈H but ∄k∈Z∣kz_0 =x) So x_0 is in H (i.e. x_0 =Min H∩R_+ ^∗ ) Let us now let us proove that H is cyclic. let x∈H. Let n be such that: n∣x_0 ∣≤∣x∣<(n+1)∣x_0 ∣ Let y:=nx_0 −x∈H By definition of x_0 it is not possible that ∣y∣∈]0,∣x_0 ∣[ It is, by definition of n impossible that ∣y∣∈[x_0 ,∞[ hence y=0 i.e. n∣x_0 ∣=∣x∣ i.e. ±nx_0 =x hence ∃n∈Z∣x=nx_0 Wich concludes the proof_□

Sorry but I think you forgot some extra hypothesis \dots

In its current form, your problem {can}^{'} t be solved .

It is very easy to see so by considering :

H = (R, +)

Its intersection with [- 1, 1] D O E S contain a

non - zero element (say \frac{π}{4} or \frac{1}{2^{n}} with n your

favorite integrer) however R is not cyclic .

To proove it consider x_{0} the element that would

generate it .

Then \forall x \in R \exists n \in Z ∣ x = {n x}_{0}

But ∄ n \in Z ∣ {n x}_{0} = \frac{x_{0}}{2} \in R

There are several possible alternative questions

you might have wanted to ask :

1) H \cap [- 1, 1] contains n o non - zero element

2) H \cap [- 1, 1] contains finitely many elements

Here are quick proofs of the two alternatives

let x_{0} = Inf H \cap R_{+}^{*}

W ether in case 1) or 2) if x_{0} is n o t in H, then by

definition of Inf, we can construct a sequence

y_{n} \in H ∣ y_{n} \underset{n \to \infty}{\to} x_{0}

and y_{n} - y_{n + 1} \to 0

Meaning H cannot be cyclic (supose it is generated

by z_{0} then \forall x \in H \exists k \in Z ∣ {k z}_{0} = x but the

taking n such that 0 <∣ y_{n} - y_{n + 1} ∣<∣ z_{0} ∣

y_{n} - y_{n + 1} \in H but ∄ k \in Z ∣ {k z}_{0} = x)

So x_{0} is in H (i . e . x_{0} = Min H \cap R_{+}^{*})

Let us now let us proove that H is cyclic .

let x \in H .

Let n be such that :

n ∣ x_{0} ∣⩽∣ x ∣< (n + 1) ∣ x_{0} ∣

Let y := {n x}_{0} - x \in H

By definition of x_{0} it is not possible that

∣ y ∣\in] 0, ∣ x_{0} ∣ [

It is, by definition of n impossible that

∣ y ∣\in [x_{0}, \infty [

hence y = 0

i . e . n ∣ x_{0} ∣=∣ x ∣

i . e . \pm {n x}_{0} = x

hence \exists n \in Z ∣ x = {n x}_{0}

Wich concludes the {proof}_{}

Commented by TheHoneyCat last updated on 28/Jun/23

just realized I made a typo. The reason why Inf=Min is not that it contradicts cyclicity but that it contradicts minimality... Sorry, I should have re-read myself.��

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