Q: Prove that the property of an operation $*$ is commutative is indeed a structural property of a binary structure < S, * >.
My attempt:
Need to show: $\Phi(a * b) = \Phi(b * a)$.
Let $\Phi: S \to S'$ be an isomorphic mapping from the binary structure $\langle S, * \rangle$ to the binary structure $\langle S', *' \rangle$.
Suppose $*$ is commutative.
Let $a,b \in S$ and $a', b' \in S'$.
Then, $\Phi(a) = a'$ and $\Phi(b) = b'$.
From that,
\begin{align*} \text{LHS} &= \Phi(a * b) = \Phi(a) *' \Phi(b) = a' *' b'\\ \text{RHS} &= \Phi(b * a) = \Phi(b) *' \Phi(a) = b' *' a' \end{align*}
Issue faced:
The issue I am facing deals with the last line from "My attempt" where the textbook has the answer:
$\Phi(b * a) = \Phi(b) *' \Phi(a) = b' * a'$
It makes sense to me that if I switch $b'$ and $a'$ in that last portion, then the $*'$ appears and it would equal $a' *' b'$, which is the LHS. However, I do not understand how the $*'$ goes directly down to $*$ without changing any orders. I thought the $*'$ would stay throughout like what happened on the LHS.
If I could get any assistance in explaining why this is the case, I would greatly appreciate it. Thank you.