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 $\order H$.

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 $\index G H$ left cosets, and therefore:
 * $\order G = \index G H \order H$

Let $G$ be of finite order.

All three numbers are finite, and the result follows.

Now let $G$ be of infinite order.

If $\index G H = n$ is finite, then $\order G = n \order H$ and so $H$ is of infinite order.

If $H$ is of finite order such that $\order H = n$, then $\order G = \index G H \times n$ and so $\index G H$ is infinite.