One-Step Subgroup Test using Subset Product

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Theorem

Let $G$ be a group.

Let $\varnothing \subset H \subseteq G$ be a non-empty subset of $G$.


Then $H$ is a subgroup of $G$ if and only if:

$H H^{-1} \subseteq H$

where:

$H^{-1}$ is the inverse of $H$
$H H ^{-1}$ is the product of $H$ with $H^{-1}$.


Proof

This is a reformulation of the One-Step Subgroup Test in terms of subset product.


Necessary Condition

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

Let $x, y \in H$.

Then by the definition of subset product:

$x y^{-1} \in H H^{-1}$


As $H \le G$, from the One-Step Subgroup Test, $x y^{-1} \in H$.

Thus $H H^{-1} \subseteq H$.

$\Box$


Sufficient Condition

Let:

$H H^{-1} \subseteq H$

From the definition of subset product:

$\forall x, y \in H: x y^{-1} \in H$

So by the One-Step Subgroup Test, $H$ is a subgroup of $G$.

$\blacksquare$


Also presented as

This result can also be presented as:

$H$ is a subgroup of $G$ if and only if:

$H^{-1} H \subseteq H$

and the same argument applies.


Sources