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Problem: Let $\alpha$ and $\beta$ be elements of $S_n$. Define the relation $\sim$ on $S_n$ by $\alpha \sim\beta$ if there exists $\sigma\in S_n$ such that $\sigma\alpha\sigma^{-1}$ = $\beta$. Prove that $\sim$ is an equivalence relation on $S_n$.

I know in order to show something is an equivalence relation, it must have all three of these properties: reflexive (if $a$, then $a=a$), symmetric (if $a=b$, then $b=a$), and transitive (if $a=b$ and $b=c$, then $a=c$).

I'm not sure how to start this proof. I don't understand how we could even show the reflexive property if there is 3 elements of $S_n$: $\alpha$, $\beta$, and $\sigma$.

I have never worked with equivalence relations so any help will appreciated.

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  • $\begingroup$ The correct definition of the relation is there exists $\sigma$ such that $\sigma\alpha\sigma^{-1}=\beta$. $\endgroup$ Oct 11 '18 at 2:08
  • $\begingroup$ I think, $k$-cycles will partition the set into classes. $\endgroup$ Oct 11 '18 at 2:28
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Reflexive: Say $\alpha \in S_n$.

$\alpha \sim \alpha$ iff there exists a $\sigma$ such that $\sigma\alpha \sigma^{-1} = \alpha$. Take $\sigma$ to be the identity permutation. Then you know there does exist a $\sigma$ such that $\sigma\alpha \sigma^{-1} = \alpha$. Therefore, $\alpha \sim \alpha$.

Just start with the actual definition of the relation and show that it holds. Hopefully this clarifies things a little, and shows how to prove the other properties.

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Someone have already shown how to show the reflexivity.

For symmetric property: if there exists a $\sigma$, such that, $\sigma \alpha \sigma^{-1}= \beta$, then, $$\sigma^{-1}(\sigma \alpha \sigma^{-1})\sigma=\sigma^{-1}\beta \sigma\implies \alpha=\sigma^{-1}\beta (\sigma^{-1})^{-1}\implies \beta\sim \alpha$$

Similarly, for transitivity, if $\alpha\sim \beta$ and $\beta\sim \gamma$, then, there exists such $\sigma_1$and $\sigma_2$ resp. Then, $$\gamma=\sigma_2\beta \sigma_2^{-1}=\sigma_2(\sigma_1\alpha\sigma_1^{-1})\sigma_2^{-1}=^{\ast}(\sigma_2\sigma_1)\alpha(\sigma_1^{-1}\sigma_2^{-1})~~~^{\ast}\text{due to associative prop.} $$now it is easy to show that $\sigma_1^{-1}\sigma_2^{-1}=(\sigma_2\sigma_1)^{-1}$. Hence, $\alpha\sim\gamma$ and we are done.

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