Subset of Abelian Group Generated by Product of Element with Inverse Element is Subgroup

{rename|Something less unwieldy which may adopt the language of Definition:Inverse Completion}}

Theorem
Let $\struct {G, \circ}$ be an abelian group.

Let $S \subset G$ be a non-empty subset of $G$ such that $\struct {S, \circ}$ is closed.

Let $H$ be the set defined as:
 * $H := \set {x \circ y^{-1}: x, y \in S}$

Then $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$.

Proof
Let $x \in S$.

Then:
 * $x \circ x^{-1} \in H$

and so $H \ne \O$.

Now let $a, b \in H$.

Then:
 * $a = x_a \circ y_a^{-1}$

and:
 * $b = x_b \circ y_b^{-1}$

for some $x_a, y_a, x_b, y_b \in S$.

Thus:

As $\struct {S, \circ}$ is closed, both $x_a \circ y_b \in S$ and $x_b \circ y_a \in S$.

Thus $a \circ b$ is in the form $x \circ y^{-1}$ for $x, y \in S$.

Thus $a \circ b \in H$

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

Also see

 * Definition:Inverse Completion