# 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:

 $\ds x^r$ $=$ $\ds x^s$ $\ds \leadsto \ \$ $\ds x^r \circ e$ $=$ $\ds x^r \circ x^{s - r}$ Definition of Identity Element, Powers of Group Elements: Sum of Indices $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds e$ $=$ $\ds x^{s - r}$ Cancellation Laws $\ds \leadsto \ \$ $\ds e$ $\in$ $\ds H$ as $H$ is closed under $\circ$

Then we have:

 $\ds e$ $=$ $\ds x^{s - r}$ which is $(1)$ $\ds \leadsto \ \$ $\ds e$ $=$ $\ds x \circ x^{s - r - 1}$ as $H$ is closed under $\circ$ $\ds \leadsto \ \$ $\ds x^{-1} \circ e$ $=$ $\ds x^{-1} \circ x \circ x^{s - r - 1}$ $\text {(2)}: \quad$ $\ds \leadsto \ \$ $\ds x^{-1}$ $=$ $\ds x^{s - r - 1}$ Definition of Inverse Element, Definition of Identity Element

But we have that:

 $\ds r$ $<$ $\ds s$ by hypothesis $\ds \leadsto \ \$ $\ds s - r$ $<$ $\ds 0$ $\ds \leadsto \ \$ $\ds s - r - 1$ $\le$ $\ds 0$ $\ds \leadsto \ \$ $\ds x^{s - r - 1}$ $\in$ $\ds \set {e, x, x^2, x^3, \ldots}$ $\ds \leadsto \ \$ $\ds x^{s - r - 1}$ $\in$ $\ds H$ as all elements of $\set {e, x, x^2, x^3, \ldots}$ are in $H$

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 $G$.

$\blacksquare$