# Finite Subgroup Test/Proof 2

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## 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 $G$.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 36.5$: Subgroups - 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.9$: Subgroups: Lemma $7$