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 $\phi^{-1}$.

$\blacksquare$


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