Lagrange's Theorem (Group Theory)/Proof 3
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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