If there is a bijection $f: X\rightarrow Y$, prove that there exists an isomorphism $\phi :S_X\rightarrow S_Y$

Question

If there is a bijection $f: X\rightarrow Y$, prove that there exists an isomorphism $\phi :S_X\rightarrow S_Y$.

Here $S_X$ denotes the group of all permutations of $X$, i.e., the bijections $X\to X$ and the group operation is composition.

I know that $X$ and $Y$ have the same cardinality because of the bijection. I define $\phi:S_X \rightarrow S_Y$ by $\phi : a \rightarrow f\circ a\circ f^{-1}$ Then I suppose $\phi^{-1}$ can be defined as $b\rightarrow f^{-1} \circ b\circ f$.

I think this is close but I haven't showed that $\phi$ is a homomorphism. But once I do I can show isomorphism.

Answer 1

$\phi(a\circ b)= f\circ (a\circ b)\circ f^{-1}=(f\circ a\circ f^{-1})\circ (f \circ b\circ f^{-1})=\phi(a)\circ \phi(b)$

Answer 2

1. What is a permutation? It is bijection from a set to itself.

2. The composition of two bijection functions is bijection.