Show that $(S,\ast)$ is commutative if and only if $(T,\Box)$ is commutative. Assume that $(S,\ast)$ and $(T,\Box)$ are isomorphic binary structures.

Show that $(S,\ast)$ is commutative if and only if $(T,\Box)$ is commutative.

Since $(S,\ast) \cong (T,\Box)$, we have $f(a \ast b) = f(a) \Box f(b)$.
I know I need to show that $a \ast b = b \ast a$ for all $a,b \in S \Longleftrightarrow a \Box b = b \Box a$ for all $a,b \in T$ but I am having trouble showing that the commutativity of $(S,\ast)$ implies that $(T,\Box)$ is also commutative. If I could get some help on one direction ($a \ast b = b \ast a$ for all $a,b \in S \Rightarrow a \Box b = b \Box a$ for all $a,b \in T$) then I could probably figure out the rest.
 A: You're on the right track.
Suppose $(S, \ast)$ is commutative and take arbitrary $a, b \in T$. Since $f: S \to T$ is an isomorphism and in particular a bijection, $a$ and $b$ have unique preimages in $S$, and by hypothesis these commute:
$$f^{-1}(a) \ast f^{-1}(b) = f^{-1}(b) \ast f^{-1}(a).$$
Now, apply $f$ to both sides.

Using that $f$ is an isomorphism (of the binary structure) gives precisely that $a, b$ commute. Since $a, b \in T$ are arbitrary, $(T, \square)$ is commutative.

A: Hint:


*

*Indeed any isomorphism $f$ has to satisfy $f(a*b) = f(a)\square f(b)$. However, isomorphisms are much stronger, in particular it has to be a bijection and $f^{-1}(a\square b) = f^{-1}(a)*f^{-1}(b)$.


I hope this helps $\ddot\smile$
A: Suppose that $(S,\ast)$ is commutative and take $a,b \in T$. Since $f:(S,\ast) \to (T,\Box)$ is an isomorphism, it follows that $f^{-1}:(T,\Box) \to (S,\ast)$ is also. Thus  $$\begin{align} f^{-1}(a \Box b) &= f^{-1}(a)\ast f^{-1}(b) \\&= f^{-1}(b) \ast f^{-1}(a) \\&= f^{-1}(b \Box a).\end{align}$$
Applying $f$ to both sides of $f^{-1}(a \Box b) = f^{-1}(b \Box a)$ yields $a \Box b = b \Box a$, demonstrating that $(T,\Box)$ is commutative.
Now assume that $(T,\Box)$ is commutative and let $a,b \in S$. Since $f$ is an isomorphism, we have $$\begin{align} f(a \ast b) &= f(a) \Box f(b) \\&=f(b) \Box f(a) \\&=f(b \ast a).\end{align}$$
We may apply the inverse $f^{-1}$ to both sides to obtain $a \ast b = b \ast a$, demonstrating that $(S, \ast)$ is commutative, as desired. $\Box$
