[[Subgroup]]
# Lagrange's theorem

Given a group $G$ and subgroup $H \sube G$, the order of the subgroup divides the order of the group. #m/thm/group 
This is often stated as
$$
\begin{align*}
\abs{G} = \abs{G / H} \abs{H}
\end{align*}
$$
where $\abs{G / H}$ is the number of unique (left or right) [[coset|cosets]] of $H$,
and is called the **Lagrange index**.

> [!check]- Proof
> Let $H \sube G$.
> Any element $g \in G$ is contained at least in the [[coset]] $gH$.
> Since [[Cosets are either identical or disjoint]],
> cosets form a [[Partition]] of $G$.
> Since $G$ is finite there is a finite number of cosets in the partition $(G : H)$.
> The number of elements in each coset is equal to $\abs{H}$.
> Therefore, $\abs{G} = (G : H) \abs{H}$.
> <span class="QED"/>

## Corollary

The order $\abs{g}$ of an element $g \in G$ divides the order $\abs{G}$ of a finite group $G$,
since $\langle g \rangle$ forms a [[Cyclic subgroup]].

## Consequences

- [[Group of prime order|All prime-ordered groups are cyclic]]

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