Finite Subgroup Test/Proof 2

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

Then:
 * $\forall a, b \in H: a \circ b \in H$

by definition of subgroup.

Necessary Condition
Let $H$ be a non-empty finite subset of $G$ such that:
 * $\forall a, b \in H: a \circ b \in H$

Let $x \in H$.

We have by hypothesis that $H$ is closed under $\circ$.

Thus all elements of $\set {x, x^2, x^3, \ldots}$ are in $H$.

But $H$ is finite.

Therefore it must be the case that:
 * $\exists r, s \in \N: x^r = x^s$

for $r < s$.

So we can write:

Then we have:

But we have that:

So from $(2)$:
 * $x^{-1} = x^{s - r - 1}$

it follows that:
 * $x^{-1} \in H$

and from the Two-Step Subgroup Test it follows that $H$ is a subgroup of $G$.