Is it an isomorphism between $S(A)$ and $S(B)$?

Question

The Problem is: $A$ and $B$ are two sets such that there is a bijection from $A$ onto $B$, and $S(A)$ and $S(B)$ denote the set of all permutations of $A$ and $B$ ten we have to prove that there is an isomorphism from $S(A)$ onto $S(B)$.

My attempt:
Let $f$ be the bijection from A onto B and $g \in S(A) $ and $h \in S(B)$ , now I define a map $\phi=h \circ f \circ g$, then later prove that this $\phi$ is well defined, bijective and is operation-preserving to show that this is an isomorphism between the two permutation groups,
but when I checked my solution on net, there seems to have a different solution for this problem, Can someone please tell me what is wrong with my solution ?

Answer 1

Your map $\phi$ isn't a map from $S(A)$ to $S(B)$. Rather, it is a map $A \to B$.

Also, you didn't define a single map $\phi$, but a whole lot of them: One for each $g \in S(A)$ and $h \in S(B)$.

To fix this, try to define a map $\phi : S(A) \to S(B)$ by declaring, for any $g \in S(A)$, which element of $S(B)$ the image $\phi(g)$ should be. Your description of $\phi(g)$ should depend on $g$, but not on any $h \in S(B)$; in fact, no $h \in S(B)$ should be in scope. Similarly, for your definition of $\phi$ no $g \in S(A)$ should be in scope and in particular $\phi$ should not depend on any $g \in S(A)$ (the values of $\phi$ may and should of course depend on the argument; but note that, generally for functions, $f(x)$ and $f$ are totally different things, so these two statement don't contradict each other).

Answer 2

Indeed, as Ingo pointed out, given

\begin{align*} f: A &\to B,\\ g: A &\to A \quad (\text{in }S(A)),\\ h: B &\to B \quad (\text{in }S(B)), \end{align*}

the composition $h \circ f \circ g$ follows the path $A \overset{g}{\to} A \overset{f}{\to} B \overset{h}{\to} B$, and is a map from $A$ to $B$, not from $S(A)$ to $S(B)$.

In order to show $S(A) \cong S(B)$, you'll need to generate a permutation $\phi(g): B \to B$ for each permutation $g: A \to A$. Thus, given a permutation $g$ of $A$, the map $\phi(g)$ will permute $B$, so $\phi$ really will be a map from $S(A)$ to $S(B)$.

You'll use only some bijection $f: A \to B$ and the group $S(A)$, the group $S(A)$ being what gives the map(s) $g$.

You've got the right idea (function composition) with $h \circ f \circ g$, but you need to start in $B$, find a way to get into $A$, permute things in $A$, then go back to $B$. Pictorially, you want maps like

$$\phi(g): B \overset{?_1}{\to} A \overset{g}{\to} A \overset{?_2}{\to}B,$$

which should be something in $S(B)$, and the composition will let you define $\phi : S(A) \to S(B)$.