Lagrange's Theorem (Group Theory)

Theorem
Let $G$ be a group of finite order.

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

Then $\left|{H}\right|$ divides $\left|{G}\right|$.

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

where:
 * $\left|{G}\right|$ and $\left|{H}\right|$ are the order of $G$ and $H$ respectively
 * $\left[{G : H}\right]$ is the index of $H$ in $G$.

When $\left|{G}\right|$ is infinite, we can still interpret this theorem sensibly:


 * A subgroup of finite index in a group of infinite order is itself of infinite order


 * A finite subgroup of a group of infinite order 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 Camille Jordan. Lagrange's proof merely showed that a subgroup of the symmetric group $S_n$ has order dividing $n!$.