Assume I have 2 sets $E$ and $F$ and there exists a bijection $\beta: E \to F$. I'm trying to show that an isomorphism exists between $S_E$ and $S_F$.
Because of $\beta$ we must have $|E| = |F|$. I think we could build $\phi: S_E \to S_F$ using $\beta$ but I'm a bit stuck on how to start from here.
$\forall e \in E, \beta(e) = f \in F$ where $f$ is unique. We also have the opposite (going from $S_F$ to $S_E$). $\beta$ has an inverse, $\beta^{-1}$ I'm sure this would come in handy in finding an isomorphism as well.
My confusion comes from what is happening to $\beta$ after a permutation on the elements of $E$.
Any tips on the best way to tackle this problem? I find myself having issues showing that 2 groups are isomorphic in general.
Thanks
The elements of $S_{E}$ are the bijections from $E$ to $E$. We can use the fact that compositions of bijective functions are bijective. In particular, we could define a map $\phi: S_{E}\rightarrow S_{F}$ by $\sigma\in S_{E}\mapsto \beta\circ\sigma\circ\beta^{-1}$.
To show this is a homomorphism, let $\sigma_{1},\sigma_{2}\in S_{E}$. Then $\phi(\sigma_{1}\circ\sigma_{2}) = \beta \circ (\sigma_{1}\circ\sigma_{2}) \circ \beta^{-1} = (\beta \circ \sigma_{1} \circ \beta^{-1}) \circ (\beta \circ \sigma_{2} \circ \beta^{-1}) = \phi(\sigma_{1})\circ\phi(\sigma_{2})$.
It should also be straightforward to show $\phi$ is bijective since $\beta$ is.
So $\phi$ will be an isomorphism as desired.