# Symmetric groups of sets with the same cardinality are isomorphic

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.

• What is the question? $\psi$ is defined correctly. – Mark May 23 at 13:48
• 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.) – Ethan Bolker May 23 at 13:49
• Sorry, let me change this from a statement to a question. – Sam.S May 23 at 13:49

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.
• It's a map $f$ from a group $G$ into a group $H$ such that$$(\forall x,y\in G):f(xy)=f(x)f(y).$$ – José Carlos Santos May 23 at 13:55