Lagrange's Theorem (Group Theory)/Proof 3

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Let $G$ be a finite group.

Let $H$ be a subgroup of $G$.


$\order H$ divides $\order G$

where $\order G$ and $\order H$ are the order of $G$ and $H$ respectively.

In fact:

$\index G H = \dfrac {\order G} {\order H}$

where $\index G H$ is the index of $H$ in $G$.

When $G$ is an infinite group, we can still interpret this theorem sensibly:

A subgroup of finite index in an infinite group is itself an infinite group.
A finite subgroup of an infinite group has infinite index.


Follows directly from the Orbit-Stabilizer Theorem applied to Group Action on Coset Space.


Source of Name

This entry was named for Joseph Louis Lagrange.