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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.

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    $\begingroup$ By $S_X$ you mean the group of permutations over $X$?all bijections from $X $onto itself? $\endgroup$
    – BigM
    Feb 20, 2015 at 5:01
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    $\begingroup$ @BigM Yes that's what I meant. $\endgroup$
    – grayQuant
    Feb 20, 2015 at 5:46

2 Answers 2

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

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  • $\begingroup$ Why $f\circ (a\circ b)\circ f^{-1}=(f\circ a\circ f^{-1})\circ (f \circ b\circ f^{-1})$? $\endgroup$
    – Wang Kah Lun
    Aug 12, 2015 at 3:59
  • $\begingroup$ Because $f$ is a bijection. $\endgroup$
    – BigM
    Aug 13, 2015 at 5:13
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1. What is a permutation? It is bijection from a set to itself.

2. The composition of two bijection functions is bijection.

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  • $\begingroup$ Your definition of permutation is not correct. It needs to be bijective. $\endgroup$
    – Tobias Kildetoft
    Feb 20, 2015 at 8:24
  • $\begingroup$ @TobiasKildetoft Usually $S_X$ denotes the symmetric group on a finite set X. The injection from a finite set to itself is bijection. :) $\endgroup$
    – user80225
    Feb 20, 2015 at 8:36
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    $\begingroup$ I disagree with the "usually". $\endgroup$
    – Tobias Kildetoft
    Feb 20, 2015 at 8:37
  • $\begingroup$ @TobiasKildetoft +1 for disagree. Anyway, I will change it. Thanks $\endgroup$
    – user80225
    Feb 20, 2015 at 8:38
  • $\begingroup$ @user31009 In a group, the elements have inverses. So how do you invert injective but non-surjective maps on an infinite set? $\endgroup$
    – Christoph
    Feb 20, 2015 at 8:46

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