Lagrange's Theorem (Group Theory)

Theorem
Let $G$ be a finite group.

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

Then:
 * $\left\vert{H}\right\vert$ divides $\left\vert{G}\right\vert$

where $\left\vert{G}\right\vert$ and $\left\vert{H}\right\vert$ are the order of $G$ and $H$ respectively.

In fact:
 * $\left[{G : H}\right] = \dfrac {\left\vert{G}\right\vert} {\left|{H}\right|}$

where $\left[{G : H}\right]$ 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.

Remark
The converse of Lagrange's Theorem is not true in general.

We consider the symmetric group $S_4$. Then the order of the alternating group $A_4$ is $12$. Now $6$ divides $12$. But there is no subgroup of order $6$.

This result, however, was actually due to. 's proof merely showed that a subgroup of the symmetric group $S_n$ has order dividing $n!$ Gosh!