Subset of Abelian Group Generated by Product of Element with Inverse Element is Subgroup
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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
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $5$: Subgroups: Exercise $3$