# 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:

 $\displaystyle a \circ b^{-1}$ $=$ $\displaystyle \paren {x_a \circ y_a^{-1} } \circ \paren {x_b \circ y_b^{-1} }^{-1}$ $\displaystyle$ $=$ $\displaystyle \paren {x_a \circ y_a^{-1} } \circ \paren {\paren {y_b^{-1} }^{-1} \circ x_b^{-1} }$ Inverse of Group Product $\displaystyle$ $=$ $\displaystyle \paren {x_a \circ y_a^{-1} } \circ \paren {y_b \circ x_b^{-1} }$ Inverse of Group Inverse $\displaystyle$ $=$ $\displaystyle x_a \circ \paren {y_a^{-1} \circ y_b} \circ x_b^{-1}$ Group Axiom $\text G 1$: Associativity $\displaystyle$ $=$ $\displaystyle x_a \circ \paren {y_b\circ y_a^{-1} } \circ x_b^{-1}$ Definition of Abelian Group $\displaystyle$ $=$ $\displaystyle \paren {x_a \circ y_b} \circ \paren {y_a^{-1} \circ x_b^{-1} }$ Group Axiom $\text G 1$: Associativity $\displaystyle$ $=$ $\displaystyle \paren {x_a \circ y_b} \circ \paren {x_b \circ y_a}^{-1}$ Inverse of Group Product

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 $G$.

$\blacksquare$

## Examples

### $\struct {\Z_{\ne 0}, \times}$ in $\struct {\R_{\ne 0}, \times}$

Let $\struct {\Z_{\ne 0}, \times}$ denote the algebraic structure formed by the set of non-zero integers under multiplication.

Let $\struct {\R_{\ne 0}, \times}$ denote the algebraic structure formed by the set of non-zero real numbers under multiplication.

Let $H$ denote the set:

$H := \set {x \times y^{-1}: x, y \in \Z_{\ne 0} }$

Then $H$ is the subgroup of $\struct {\R_{\ne 0}, \times}$ which is the set of non-zero rational numbers under multiplication:

$H = \struct {\Q_{\ne 0}, \times}$