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Question Number 192094 by Mastermind last updated on 08/May/23

Prove that the order of a subgroup  S of a finite group G, always divide  the order of group G.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{a}\:\mathrm{subgroup} \\ $$$$\mathrm{S}\:\mathrm{of}\:\mathrm{a}\:\mathrm{finite}\:\mathrm{group}\:\mathrm{G},\:\mathrm{always}\:\mathrm{divide} \\ $$$$\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{group}\:\mathrm{G}. \\ $$

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