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

From ProofWiki
Jump to navigation Jump to search


It has been suggested that this page be renamed.
In particular: Something less unwieldy which may adopt the language of Definition:Inverse Completion
To discuss this page in more detail, feel free to use the talk page.


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:

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

Retrieved from "https://proofwiki.org/w/index.php?title=Subset_of_Abelian_Group_Generated_by_Product_of_Element_with_Inverse_Element_is_Subgroup&oldid=476568"