Opposite Group is Group
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Theorem
Let $\struct {G, \circ}$ be a group.
Let $\struct {G, *}$ be the opposite group to $G$.
Then $\struct {G, *}$ is a group.
Proof
Group Axiom $\text G 0$: Closure
$\struct {G, *}$ is closed:
- $b \circ a \in G \implies a * b \in G$
$\Box$
Group Axiom $\text G 1$: Associativity
$*$ is associative on $G$:
\(\ds a * \paren {b * c}\) | \(=\) | \(\ds \paren {c \circ b} \circ a\) | Definition of $*$ | |||||||||||
\(\ds \) | \(=\) | \(\ds c \circ \paren {b \circ a}\) | Associativity of $\circ$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a * b} * c\) | Definition of $*$ |
$\Box$
Group Axiom $\text G 2$: Existence of Identity Element
Let $e$ be the identity of $\struct {G, \circ}$:
\(\ds a * e\) | \(=\) | \(\ds e \circ a = a\) | ||||||||||||
\(\ds e * a\) | \(=\) | \(\ds a \circ e = a\) |
Thus $e$ is the identity of $\struct {G, *}$.
$\Box$
Group Axiom $\text G 3$: Existence of Inverse Element
Let the inverse of $a \in \struct {G, \circ}$ be $a^{-1}$:
\(\ds a * a^{-1}\) | \(=\) | \(\ds a^{-1} \circ a = e\) | ||||||||||||
\(\ds a^{-1} * a\) | \(=\) | \(\ds a \circ a^{-1} = e\) |
Thus $a^{-1}$ is the inverse of $a \in \struct {G, *}$
$\Box$
So all the group axioms are satisfied, and $\struct {G, *}$ is a group.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 26 \epsilon$