Finite Subgroup Test

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

Let $$H$$ be a 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.