Group theory and generalizations Theorem

A subgroup whose index is the smallest prime divisor of the order is normal

Builds on Normal subgroup Index of a subgroup

Statement

Let $G$ be a finite group and let $p$ be the smallest prime number that divides $|G|$ . Let $H$ be a subgroup of $G$ of index $p$ . Then $H$ is normal in $G$ .

Remarks

For $p = 2$ this is the fact that subgroups of index two are normal. The proof lets $G$ act on the $p$ left cosets of $H$ . The kernel $K$ of this action lies in $H$ , and $[G : K]$ divides both $|G|$ and $p!$ . Since $p$ is the smallest prime divisor of $|G|$ , this forces $[G : K] = p$ , so $H = K$ is normal.

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Group concept
Conjugate elements definition
Subgroup definition
Conjugate subgroup definition
Left coset definition
Normal subgroup definition
Index of a subgroup definition
A subgroup whose index is the smallest prime divisor of the order is normal statement

Tag	`0FEBE9NC0E`
URI	`m2://0FEBE9NC0E`
Kind	statement, theorem
Writing	W0 stub
MSC	20D60
Status	This copy of the site has no build data. Run `m2 status 0FEBE9NC0E`.
Uses	Normal subgroup, Index of a subgroup

A subgroup whose index is the smallest prime divisor of the order is normal

Statement

Remarks

Reading path

Further reading

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