# Finite Subgroup Test

## 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 1

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.

$\blacksquare$

## 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.

$\Box$

### 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:

\(\displaystyle x^r\) | \(=\) | \(\displaystyle x^s\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x^r \circ e\) | \(=\) | \(\displaystyle x^r \circ x^{s - r}\) | $\quad$ Definition of Identity Element, Powers of Group Elements: Sum of Indices | $\quad$ | ||||||||

\((1):\quad\) | \(\displaystyle \leadsto \ \ \) | \(\displaystyle e\) | \(=\) | \(\displaystyle x^{s - r}\) | $\quad$ Cancellation Laws | $\quad$ | |||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle e\) | \(\in\) | \(\displaystyle H\) | $\quad$ as $H$ is closed under $\circ$ | $\quad$ |

Then we have:

\(\displaystyle e\) | \(=\) | \(\displaystyle x^{s - r}\) | $\quad$ which is $(1)$ | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle e\) | \(=\) | \(\displaystyle x \circ x^{s - r - 1}\) | $\quad$ as $H$ is closed under $\circ$ | $\quad$ | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x^{-1} \circ e\) | \(=\) | \(\displaystyle x^{-1} \circ x \circ x^{s - r - 1}\) | $\quad$ | $\quad$ | ||||||||

\((2):\quad\) | \(\displaystyle \leadsto \ \ \) | \(\displaystyle x^{-1}\) | \(=\) | \(\displaystyle x^{s - r - 1}\) | $\quad$ Definition of Inverse Element, Definition of Identity Element | $\quad$ |

But we have that:

\(\displaystyle r\) | \(<\) | \(\displaystyle s\) | $\quad$ by hypothesis | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s - r\) | \(<\) | \(\displaystyle 0\) | $\quad$ | $\quad$ | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s - r - 1\) | \(\le\) | \(\displaystyle 0\) | $\quad$ | $\quad$ | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x^{s - r - 1}\) | \(\in\) | \(\displaystyle \set {e, x, x^2, x^3, \ldots}\) | $\quad$ | $\quad$ | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x^{s - r - 1}\) | \(\in\) | \(\displaystyle H\) | $\quad$ as all elements of $\set {e, x, x^2, x^3, \ldots}$ are in $H$ | $\quad$ |

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 $G$.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): Chapter $5$: Subgroups: Exercise $15$