# Lagrange's Theorem (Group Theory)

## 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 1

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

$\blacksquare$

## Proof 2

Let $G$ be a group.

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

From Cosets are Equivalent, a left coset $y H$ has the same number of elements as $H$, namely $\order H$.

Since left cosets are identical or disjoint, each element of $G$ belongs to exactly one left coset.

From the definition of index of subgroup, there are $\index G H$ left cosets, and therefore:

$\order G = \index G H \order H$

Let $G$ be of finite order.

All three numbers are finite, and the result follows.

Now let $G$ be of infinite order.

If $\index G H = n$ is finite, then $\order G = n \order H$ and so $H$ is of infinite order.

If $H$ is of finite order such that $\order H = n$, then $\order G = \index G H \times n$ and so $\index G H$ is infinite.

$\blacksquare$

## Proof 3

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

$\blacksquare$

## Examples

### Intersection of Subgroups of Order $25$ and $36$

Let $G$ be a group.

Let $H$ and $K$ be subgroups of $G$ such that:

$\order H = 25$
$\order K = 36$

where $\order {\, \cdot \,}$ denotes the order of the subgroup.

Then:

$\order {H \cap K} = 1$

### Order of Group with Subgroups of Order $25$ and $36$

Let $G$ be a group.

Let $H$ and $K$ be subgroups of $G$ such that:

$\order H = 25$
$\order K = 36$

where $\order {\, \cdot \,}$ denotes the order of the subgroup.

Then:

$900 \divides \order G$

where $\divides$ denotes divisibility.

### Order of Union of Subgroups of Order $16$

Let $G$ be a group whose identity is $e$.

Let $H$ and $K$ be subgroups of $G$ such that:

$\order H = \order K = 16$
$H \ne K$

where $\order {\, \cdot \,}$ denotes the order of the subgroup.

Then:

$24 \le \order {H \cup K} \le 31$

## Source of Name

This entry was named for Joseph Louis Lagrange.

## Historical Note

Lagrange's theorem was actually proved by Camille Jordan.

Lagrange's proof merely showed that a subgroup of the symmetric group $S_n$ has an order which is a divisor of $n!$.