Lagrange's Theorem (Group Theory)/Proof 2

Proof
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 $\left|{H}\right|$.

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.