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