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:
\(\ds x^r\) | \(=\) | \(\ds x^s\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^r \circ e\) | \(=\) | \(\ds x^r \circ x^{s - r}\) | 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
- 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 $7$