Finite Subgroup Test

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $H$ be a non-empty finite subset of $G$.

Suppose $a, b \in H \implies a \circ b \in H$.

Then $H \le G$, i.e. $H$ is a subgroup of $G$.

That is, a finite subset of $G$ is a subgroup iff it is closed.

Proof
Let $H$ be a finite subset of $G$ such that $a, b \in H \implies a \circ b \in H$.

From the Two-Step Subgroup Test, it follows that we only need to show that $a \in H \implies a^{-1} \in H$.

So, let $a \in H$.

First it is straightforward to show by induction that $\left\{{x \in G: x = a^n: n \in \N^*}\right\} \subseteq H$.

That is, $a \in H \implies \forall n \in \N: a^n \in H$.

Now, since $H$ is finite, we have that the order of $a$ is finite.

Let the order of $a$ be $m$.

From Inverse Element is Power of Order Less 1 we have that $a^{m-1} = a^{-1}$.

As $a^{m-1} \in H$ (from above) the result follows.