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

Answered by Aziztisffola last updated on 13/Aug/20

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

Answered by Aziztisffola last updated on 13/Aug/20

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

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