# Finite Subgroup Test/Proof 2

## 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

### 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:

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

Then we have:

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

But we have that:

 $\displaystyle r$ $<$ $\displaystyle s$ by hypothesis $\displaystyle \leadsto \ \$ $\displaystyle s - r$ $<$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle s - r - 1$ $\le$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle x^{s - r - 1}$ $\in$ $\displaystyle \set {e, x, x^2, x^3, \ldots}$ $\displaystyle \leadsto \ \$ $\displaystyle x^{s - r - 1}$ $\in$ $\displaystyle H$ 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 $G$.

$\blacksquare$