Finite Subgroup Test

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

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


$H$ is a subgroup of $G$

if and only if:

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

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.


Proof 2

Sufficient Condition

Let $H$ be a subgroup of $G$.


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

by definition of subgroup.


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 $G$.