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 $G$

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

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

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