# Equivalence relation on $S_n$ involving algebraic permutations

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.

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

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, $$\alpha \sim \alpha$$.
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.