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

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{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 $G \, 1$: 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 $G \, 1$: 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:

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


Also see


Sources