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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.)

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  • $\begingroup$ I am really not sure what exactly I am expected, and proving this using cardinals seems too obvious.. $\endgroup$ – Meitar Jan 9 '15 at 12:40
  • $\begingroup$ I'm ignorant, what kind of a function is Sym? $\endgroup$ – Regret Jan 9 '15 at 12:40
  • $\begingroup$ 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... $\endgroup$ – Meitar Jan 9 '15 at 12:44
  • $\begingroup$ Are your sets finite? $\endgroup$ – Ofir Schnabel Jan 9 '15 at 12:49
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    $\begingroup$ Uh, $\aleph_0\cdot\aleph_0=\aleph_0$. You should also avoid subtracting cardinals, since that's pretty much meaningless for infinite cardinals. $\endgroup$ – Asaf Karagila♦ Jan 9 '15 at 13:48
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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.

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  • $\begingroup$ Are you sure you want the composition that way around? $\endgroup$ – HSN Jan 9 '15 at 12:54
  • $\begingroup$ Why is that a problem? HSN? I want to know... $\endgroup$ – Meitar Jan 9 '15 at 12:57
  • $\begingroup$ 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. $\endgroup$ – HSN Jan 9 '15 at 12:58
  • $\begingroup$ True, fixed it, thanks. $\endgroup$ – Ofir Schnabel Jan 9 '15 at 13:00

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