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Question Number 107925 by mohammad17 last updated on 13/Aug/20

Answered by Aziztisffola last updated on 13/Aug/20

True by lagrange theorem    The order than subgroup H of G divides   the order of G then if the order than  element h in H divides the order of H   hence ∣h∣ divides ∣G∣.

$$\mathrm{True}\:\mathrm{by}\:\mathrm{lagrange}\:\mathrm{theorem}\: \\ $$$$\:\mathrm{The}\:\mathrm{order}\:\mathrm{than}\:\mathrm{subgroup}\:\mathrm{H}\:\mathrm{of}\:\mathrm{G}\:\mathrm{divides} \\ $$$$\:\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{G}\:\mathrm{then}\:\mathrm{if}\:\mathrm{the}\:\mathrm{order}\:\mathrm{than} \\ $$$$\mathrm{element}\:\mathrm{h}\:\mathrm{in}\:\mathrm{H}\:\mathrm{divides}\:\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{H} \\ $$$$\:\mathrm{hence}\:\mid\mathrm{h}\mid\:\mathrm{divides}\:\mid\mathrm{G}\mid. \\ $$

Answered by Aziztisffola last updated on 13/Aug/20

Let G be a finite group  x∈G then ∣x∣=∣<x>∣ where <x> is   subgroup of G  ⇒^(Lagrange theorem)   ∣<x>∣ divides ∣G∣   ⇒∣x∣ divides ∣G∣.

$$\mathrm{Let}\:\mathrm{G}\:\mathrm{be}\:\mathrm{a}\:\mathrm{finite}\:\mathrm{group} \\ $$$$\mathrm{x}\in\mathrm{G}\:\mathrm{then}\:\mid\mathrm{x}\mid=\mid<\mathrm{x}>\mid\:\mathrm{where}\:<\mathrm{x}>\:\mathrm{is}\: \\ $$$$\mathrm{subgroup}\:\mathrm{of}\:\mathrm{G} \\ $$$$\overset{\mathrm{Lagrange}\:\mathrm{theorem}} {\Rightarrow}\:\:\mid<\mathrm{x}>\mid\:\mathrm{divides}\:\mid\mathrm{G}\mid \\ $$$$\:\Rightarrow\mid\mathrm{x}\mid\:\mathrm{divides}\:\mid\mathrm{G}\mid. \\ $$

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