Isomorphism Preserves Associativity/Proof 1
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Theorem
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ is associative if and only if $*$ is associative.
Proof
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
As an isomorphism is surjective, it follows that:
- $\forall u, v, w \in T: \exists x, y, z \in S: \map \phi x = u, \map \phi y = v, \map \phi z = w$
So:
\(\ds \paren {u * v} * w\) | \(=\) | \(\ds \paren {\map \phi x * \map \phi y} * \map \phi z\) | as $\phi$ is a Surjection | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {x \circ y} * \map \phi z\) | Definition of Morphism Property | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {\paren {x \circ y} \circ z}\) | Definition of Morphism Property | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {x \circ \paren {y \circ z} }\) | Associativity of $\circ$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x * \map \phi {y \circ z}\) | Definition of Morphism Property | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x * \paren {\map \phi y * \map \phi z}\) | Definition of Morphism Property | |||||||||||
\(\ds \) | \(=\) | \(\ds u * \paren {v * w}\) | by definition as above |
As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also an isomorphism.
Thus the result for $\phi$ can be applied to $\phi^{-1}$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Theorem $6.2: \ 1^\circ$