Intuitively I know that there are isomorphisms between the symmetric groups with same cardinality.
but i dont know how to show it precisely.
give me some intuition! thank you!
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Uh, you probably don't want to say "Give me....". – Eleven-Eleven Apr 19 '13 at 02:13
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@MetinY. Y.thank you! – user73309 Apr 19 '13 at 02:29
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@ChristopherErnst i dont know what you mean... I'm not english native, tell me straightforward way. – user73309 Apr 19 '13 at 02:31
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What he probably means is that this way of talking "give me..." seems as if you are demanding the answer from others. People are more likely to respond to a more polite request. – Apr 19 '13 at 03:05
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@Shahab I should look through other articles to learn how they talking:) Thanks a lot! – user73309 Apr 19 '13 at 03:17
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Let $A$ and $B$ be two sets and suppose there exists a bijection $f: A \rightarrow B$ between them. Consider the map $\phi : \text{Sym}(A) \rightarrow \text{Sym}(B)$ given by $\phi(\sigma) = f \circ \sigma \circ f^{-1}$. Prove that $\phi$ is a well-defined homomorphism and then either (1) construct an inverse homomorphism $\phi^{-1} : \text{Sym}(B) \rightarrow \text{Sym}(A)$ or (2) prove that $\phi$ is bijective. To show that $\phi$ is well-defined, you must argue why $\phi(\sigma) : B \rightarrow B$ is a bijection if $\sigma : A \rightarrow A$ is.
Michael Joyce
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