Lagrange's Theorem (Group Theory)/Proof 2

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.

Proof
For any group $G$:

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

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 $\left[{G : H}\right]$ left cosets, and therefore $\left|{G}\right| = \left[{G : H}\right] \left|{H}\right|$.

Let $G$ be of finite order.

All three numbers are finite, and the result follows.

Now let $G$ be of infinite order.

If $\left[{G : H}\right] = n$ is finite, then $\left|{G}\right| = n\left|{H}\right| \implies \left|{H}\right|$ is infinite.

If $\left|{H}\right| = n$ is finite, then $\left|{G}\right| = \left[{G : H}\right] n \implies \left[{G:H}\right]$ is infinite.