Isomorphism Preserves Commutativity/Proof 1

Theorem
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an isomorphism.

Then $\circ$ is commutative iff $*$ is commutative.

Proof
Let $\left({S, \circ}\right)$ be an algebraic structure in which $\circ$ is commutative.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an isomorphism.

As an isomorphism is surjective, it follows that:


 * $\forall u, v \in T: \exists x, y \in S: \phi \left({x}\right) = u, \phi \left({y}\right) = v$

So:

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 $\phi^{-1}$.