# Prove that if A~B then Sym(A)~Sym(B).

I tried to prove it with sets. Really, truly clumsy. I know |A|=|B|. Can I simply conclude that |A|!=|B|! => Sym(A)~Sym(B)?? (Sym(A) for a set A is the set of all bijections from A to A.)

• I am really not sure what exactly I am expected, and proving this using cardinals seems too obvious.. – Meitar Jan 9 '15 at 12:40
• I'm ignorant, what kind of a function is Sym? – Regret Jan 9 '15 at 12:40
• Sym(A) for a set A is the set of all bijections from A to A. Permutations actually... I am sorry I should have mentioned it... – Meitar Jan 9 '15 at 12:44
• Are your sets finite? – Ofir Schnabel Jan 9 '15 at 12:49
• Uh, $\aleph_0\cdot\aleph_0=\aleph_0$. You should also avoid subtracting cardinals, since that's pretty much meaningless for infinite cardinals. – Asaf Karagila Jan 9 '15 at 13:48

Take a bijection $$f:A\rightarrow B.$$ Now take a permutation $\sigma \in Sym(A)$.
Show that $f\circ\sigma\circ f^{-1}$ is a permutation of $B$.
And then show that the map $$Sym(A)\rightarrow Sym(B),$$ taking $\sigma$ to $f\circ\sigma\circ f^{-1}$ is a bijection.
• I seem to have misread it. It should probably be $f\circ\sigma\circ f^{-1}$. The function $f$ isn't defined on $B$, so $\sigma\circ f$ can't be a permutation of B. – HSN Jan 9 '15 at 12:58