# Finite Subgroup Test/Proof 1

## Theorem

Let $\struct {G, \circ}$ be a group.

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

Then:

- $H$ is a subgroup of $G$

- $\forall a, b \in H: a \circ b \in H$

That is, a non-empty finite subset of $G$ is a subgroup if and only if 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 $\set {x \in G: x = a^n: n \in \N} \subseteq H$.

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.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 35$ - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $4$: Subgroups: Proposition $4.2$: Remark $4$