Isomorphism Preserves Commutativity/Proof 1

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 commutative if and only if $*$ is commutative.

Proof

Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.

As an isomorphism is surjective, it follows that:

$\forall u, v \in T: \exists x, y \in S: \map \phi x = u, \map \phi y = v$

So:

\(\ds u * v\)

\(=\)

\(\ds \map \phi x * \map \phi y\)

as $\phi$ is a surjection

\(\ds \)

\(=\)

\(\ds \map \phi {x \circ y}\)

Morphism Property

\(\ds \)

\(=\)

\(\ds \map \phi {y \circ x}\)

Commutativity of $\circ$

\(\ds \)

\(=\)

\(\ds \map \phi y * \map \phi x\)

Morphism Property

\(\ds \)

\(=\)

\(\ds v * u\)

by definition as above

As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism.

Thus the result for $\phi$ can be applied to $\phi^{-1}$.

$\blacksquare$