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 $$G$$.

Proof
Let $$H$$ be a finite subset of $$G$$.

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 $$\left\{{x \in G: x = a^n: n \in \N^*}\right\} \subseteq H$$.

Now, since $$H$$ is finite, we have from Finite Semigroup Equal Elements for Different Powers that $$a^n = a^m$$ for some $$m, n \in \N^*, n > m$$.

So, $$a^{-m} \in G \implies a^{n-m} = a^n \circ a^{-m} = a^m \circ a^{-m} = e$$.

Either:
 * $$n = m + 1 \implies a = a^{n-m} = e$$

or:
 * $$n > m + 1 \implies \exists k \in \N^*, k > 1: a^k = a^{n-m} = e$$

So either $$a^{-1} = e = a \in H$$ or $$a^{-1} = a^{k-1} \in H$$.