Lagrange's Theorem (Group Theory)/Proof 3

Theorem

Let $G$ be a finite group.

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

Then:

$\order H$ 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.

Proof

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

$\blacksquare$

Source of Name

This entry was named for Joseph Louis Lagrange.

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Theorem $10.16$: Remark