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Let $X$ and $Y$ be two sets s.t. $|X|=|Y|.$ Show that the groups $\operatorname{Sym}(X)$ and $\operatorname{Sym}(Y)$ of all permutations of $X$ and $Y$, respectively, are isomorphic.

My attempt: Since $|X|=|Y|$ we can assume there exists a bijection $\phi:X\rightarrow{Y}$. I remember successfully trying this question before with success by defining $\psi:\operatorname{Sym}(X)\rightarrow{\operatorname{Sym}(Y)},$ defined by $\psi(\sigma):=\phi{}\circ\sigma\circ\phi^{-1}$, but can't quite remember its use.

Note: I am trying to do this with basic assumed knowledge on abstract algebra. Also I am aware there are two other questions that have answers with very similar questions, but I would like a more simplistic approach.

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  • $\begingroup$ What is the question? $\psi$ is defined correctly. $\endgroup$ – Mark May 23 at 13:48
  • $\begingroup$ What is there to remember? Just show that the $\psi$ you defined is an isomorphism. (Note: your hypothesis is that the cardinality of the sets that's the same, not the cardinality of the groups. That can be tricky when the sets are infinite.) $\endgroup$ – Ethan Bolker May 23 at 13:49
  • $\begingroup$ Sorry, let me change this from a statement to a question. $\endgroup$ – Sam.S May 23 at 13:49
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The map $\psi$ is a map from $\operatorname{Sym}(X)$ into $\operatorname{Sym}(Y)$. It is easy to check that it is a group homomorphism. Besides, if you define $\eta\colon\operatorname{Sym}(Y)\longrightarrow\operatorname{Sym}(X)$ by $\eta(\sigma)=\phi^{-1}\circ\sigma\circ\phi$, then $\eta$ is clarly the inverse of $\psi$, besides being a group homomorphism too. Therefore, $\psi$ is actually a group isomorphism.

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  • $\begingroup$ Whats the difference between a homomorphism and isomorphism? $\endgroup$ – Sam.S May 23 at 13:53
  • $\begingroup$ It's a map $f$ from a group $G$ into a group $H$ such that$$(\forall x,y\in G):f(xy)=f(x)f(y).$$ $\endgroup$ – José Carlos Santos May 23 at 13:55
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    $\begingroup$ An isomorphism is a homomorphism which is also a bijection. $\endgroup$ – Mark May 23 at 14:00

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