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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.

LetHbeasubgroupof(R,+)suchthatH∩[−1,1]containsanonzeroelement.ProvethatHiscyclic.

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_□

SorrybutIthinkyouforgotsomeextrahypothesis...Initscurrentform,yourproblemcan′tbesolved.Itisveryeasytoseesobyconsidering:H=(R,+)Itsintersectionwith[−1,1]DOEScontainanon−zeroelement(sayπ4or12nwithnyourfavoriteintegrer)howeverRisnotcyclic.Toprooveitconsiderx0theelementthatwouldgenerateit.Then∀x∈R∃n∈Z∣x=nx0But∄n∈Z∣nx0=x02∈RThereareseveralpossiblealternativequestionsyoumighthavewantedtoask:1)H∩[−1,1]containsnonon−zeroelement2)H∩[−1,1]containsfinitelymanyelementsHerearequickproofsofthetwoalternativesletx0=InfH∩R+∗Wetherincase1)or2)ifx0isnotinH,thenbydefinitionofInf,wecanconstructasequenceyn∈H∣yn→n→∞x0andyn−yn+1→0MeaningHcannotbecyclic(suposeitisgeneratedbyz0then∀x∈H∃k∈Z∣kz0=xbutthetakingnsuchthat0<∣yn−yn+1∣<∣z0∣yn−yn+1∈Hbut∄k∈Z∣kz0=x)Sox0isinH(i.e.x0=MinH∩R+∗)LetusnowletusproovethatHiscyclic.letx∈H.Letnbesuchthat:n∣x0∣⩽∣x∣<(n+1)∣x0∣Lety:=nx0−x∈HBydefinitionofx0itisnotpossiblethat∣y∣∈]0,∣x0∣[Itis,bydefinitionofnimpossiblethat∣y∣∈[x0,∞[hencey=0i.e.n∣x0∣=∣x∣i.e.±nx0=xhence∃n∈Z∣x=nx0Wichconcludestheproof◻

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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