Lagrange's Theorem (Group Theory)/Proof 1

From ProofWiki
Jump to navigation Jump to search


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.


Let $G$ be finite.

Consider the mapping $\phi: G \to G / H^l$, defined as:

$\phi: G \to G / H^l: \map \phi x = x H^l$

where $G / H^l$ is the left coset space of $G$ modulo $H$.

For every $y H \in G / H^l$, there exists a corresponding $y \in G$, so $\phi$ is a surjection.

From Cardinality of Surjection it follows that $G / H^l$ is finite.

From Cosets are Equivalent, $G / H^l$ has the same number of elements as $H$.

We have that the $G / H^l$ is a partition of $G$.

It follows from Number of Elements in Partition that $\index G H = \dfrac {\order G} {\order H}$


Source of Name

This entry was named for Joseph Louis Lagrange.