Find isomorphism between $S_E$ and $S_F$ if a bijection $E\to F$ exists

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

• What could $S_E,S_F$ possibly be? Oct 1, 2015 at 0:50
• As is often the case, it may be helpful to look at some small examples. Suppose $E$ and $F$ both have $2$ elements. Can you write down an isomorphism then? What if they both have $3$ elements? Oct 1, 2015 at 1:06

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.
• Let me add a small detail. The map $\phi$ is well defined beacause $\beta \circ \sigma \circ \beta^{-1}: F \rightarrow F$ for every $\sigma \in S_{E}$ and that the composition of bijections is a bijection. So, $\beta \circ \sigma \circ \beta^{-1}$ is in fact a permutation of $F$. Oct 1, 2015 at 15:19