Question Number 193794 by Rajpurohith last updated on 20/Jun/23
![Let H be a subgroup of (R,+) such that H∩[−1,1] contains a non zero element. Prove that H is cyclic.](https://www.tinkutara.com/question/Q193794.png)
$${Let}\:{H}\:{be}\:{a}\:{subgroup}\:{of}\:\left(\mathbb{R},+\right)\:{such}\:{that}\:{H}\cap\left[−\mathrm{1},\mathrm{1}\right]\: \\ $$$${contains}\:{a}\:{non}\:{zero}\:{element}. \\ $$$${Prove}\:{that}\:{H}\:{is}\:{cyclic}. \\ $$
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_□](https://www.tinkutara.com/question/Q194114.png)
$$ \\ $$$$\mathrm{Sorry}\:\mathrm{but}\:\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{forgot}\:\mathrm{some}\:\mathrm{extra}\:\mathrm{hypothesis}… \\ $$$$\mathrm{In}\:\mathrm{its}\:\mathrm{current}\:\mathrm{form},\:\mathrm{your}\:\mathrm{problem}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{solved}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{very}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{so}\:\mathrm{by}\:\mathrm{considering}: \\ $$$${H}=\left(\mathbb{R},+\right) \\ $$$$\mathrm{Its}\:\mathrm{intersection}\:\mathrm{with}\:\left[−\mathrm{1},\mathrm{1}\right]\:{DOES}\:\mathrm{contain}\:\mathrm{a}\: \\ $$$$\mathrm{non}\:−\mathrm{zero}\:\mathrm{element}\:\left(\mathrm{say}\:\frac{\pi}{\mathrm{4}}\:\mathrm{or}\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\mathrm{with}\:{n}\:\mathrm{your}\right. \\ $$$$\left.\mathrm{favorite}\:\mathrm{integrer}\right)\:\mathrm{however}\:\mathbb{R}\:\mathrm{is}\:\mathrm{not}\:\mathrm{cyclic}. \\ $$$$\mathrm{To}\:\mathrm{proove}\:\mathrm{it}\:\mathrm{consider}\:{x}_{\mathrm{0}} \:\mathrm{the}\:\mathrm{element}\:\mathrm{that}\:\mathrm{would} \\ $$$$\mathrm{generate}\:\mathrm{it}. \\ $$$$\mathrm{Then}\:\forall{x}\in\mathbb{R}\:\exists{n}\in\mathbb{Z}\mid\:{x}={nx}_{\mathrm{0}} \\ $$$$\mathrm{But}\:\nexists{n}\in\mathbb{Z}\mid\:{nx}_{\mathrm{0}} =\frac{{x}_{\mathrm{0}} }{\mathrm{2}}\in\mathbb{R} \\ $$$$\mathrm{There}\:\mathrm{are}\:\mathrm{several}\:\mathrm{possible}\:\mathrm{alternative}\:\mathrm{questions} \\ $$$$\mathrm{you}\:\mathrm{might}\:\mathrm{have}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{ask}: \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:\:{H}\cap\left[−\mathrm{1},\mathrm{1}\right]\:\mathrm{contains}\:{no}\:\mathrm{non}−\mathrm{zero}\:\mathrm{element} \\ $$$$\left.\mathrm{2}\right)\:\:{H}\cap\left[−\mathrm{1},\mathrm{1}\right]\:\mathrm{contains}\:\mathrm{finitely}\:\mathrm{many}\:\mathrm{elements} \\ $$$$ \\ $$$$\mathrm{Here}\:\mathrm{are}\:\mathrm{quick}\:\mathrm{proofs}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{alternatives} \\ $$$$\mathrm{let}\:{x}_{\mathrm{0}} =\mathrm{Inf}\:{H}\cap\mathbb{R}_{+} ^{\ast} \\ $$$$\left.\mathrm{W}\left.\mathrm{ether}\:\mathrm{in}\:\mathrm{case}\:\mathrm{1}\right)\:\mathrm{or}\:\mathrm{2}\right)\:\mathrm{if}\:{x}_{\mathrm{0}} \:\mathrm{is}\:{not}\:\mathrm{in}\:{H},\:\mathrm{then}\:\mathrm{by} \\ $$$$\mathrm{definition}\:\mathrm{of}\:\mathrm{Inf},\:\mathrm{we}\:\mathrm{can}\:\mathrm{construct}\:\mathrm{a}\:\mathrm{sequence} \\ $$$${y}_{{n}} \in{H}\:\mid\:{y}_{{n}\:\:} \underset{{n}\rightarrow\infty} {\rightarrow}{x}_{\mathrm{0}} \\ $$$$\mathrm{and}\:{y}_{{n}} −{y}_{{n}+\mathrm{1}} \rightarrow\mathrm{0} \\ $$$$\mathrm{Meaning}\:{H}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{cyclic}\:\left(\mathrm{supose}\:\mathrm{it}\:\mathrm{is}\:\mathrm{generated}\right. \\ $$$$\mathrm{by}\:{z}_{\mathrm{0}} \:\mathrm{then}\:\forall{x}\in{H}\:\exists{k}\in\mathbb{Z}\mid\:{kz}_{\mathrm{0}} ={x}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{taking}\:{n}\:\mathrm{such}\:\mathrm{that}\:\mathrm{0}<\mid{y}_{{n}} −{y}_{{n}+\mathrm{1}} \mid<\mid{z}_{\mathrm{0}} \mid \\ $$$$\left.{y}_{{n}} −{y}_{{n}+\mathrm{1}} \in{H}\:\mathrm{but}\:\nexists{k}\in\mathbb{Z}\mid{kz}_{\mathrm{0}} ={x}\right) \\ $$$$\mathrm{So}\:{x}_{\mathrm{0}} \:\mathrm{is}\:\mathrm{in}\:{H}\:\left({i}.{e}.\:{x}_{\mathrm{0}} =\mathrm{Min}\:{H}\cap\mathbb{R}_{+} ^{\ast} \right) \\ $$$$ \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{now}\:\mathrm{let}\:\mathrm{us}\:\mathrm{proove}\:\mathrm{that}\:{H}\:\mathrm{is}\:\mathrm{cyclic}. \\ $$$$\mathrm{let}\:{x}\in{H}. \\ $$$$\mathrm{Let}\:{n}\:\mathrm{be}\:\mathrm{such}\:\mathrm{that}: \\ $$$${n}\mid{x}_{\mathrm{0}} \mid\leqslant\mid{x}\mid<\left({n}+\mathrm{1}\right)\mid{x}_{\mathrm{0}} \mid \\ $$$$\mathrm{Let}\:{y}:={nx}_{\mathrm{0}} −{x}\in{H} \\ $$$$\mathrm{By}\:\mathrm{definition}\:\mathrm{of}\:{x}_{\mathrm{0}} \:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{possible}\:\mathrm{that}\: \\ $$$$\left.\mid{y}\mid\in\right]\mathrm{0},\mid{x}_{\mathrm{0}} \mid\left[\right. \\ $$$$\mathrm{It}\:\mathrm{is},\:\mathrm{by}\:\mathrm{definition}\:\mathrm{of}\:{n}\:\mathrm{impossible}\:\mathrm{that} \\ $$$$\mid{y}\mid\in\left[{x}_{\mathrm{0}} ,\infty\left[\right.\right. \\ $$$$\mathrm{hence}\:{y}=\mathrm{0} \\ $$$${i}.{e}.\:{n}\mid{x}_{\mathrm{0}} \mid=\mid{x}\mid \\ $$$${i}.{e}.\:\pm{nx}_{\mathrm{0}} ={x} \\ $$$$\mathrm{hence}\:\exists{n}\in\mathbb{Z}\mid{x}={nx}_{\mathrm{0}} \\ $$$$\mathrm{Wich}\:\mathrm{concludes}\:\mathrm{the}\:\mathrm{proof}_{\Box} \\ $$
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.��