If $α,ρ ∈ S_n$ show the fixed set of $ραρ^{−1}$ equals $ρ(\mathrm{Fix}(α))$

Question

If α,ρ ∈ Sn show the fix(ραρ−1)=ρ(fix(α))

Note: the fix(ραρ−1) ={x∈X|ραρ−1(x)=x} and the fix(α)={x∈X|α(x)=x}

TIA

Answer 1

Let's think about how to show $\text{Fix}(\rho\alpha\rho^{-1}) \subset \rho \text{Fix}(\alpha)$. First translate what it means to be a member of each set:

1. $x \in \text{Fix}(\rho\alpha\rho^{-1})$ means $x$ is fixed by $\rho\alpha\rho^{-1}$, i.e. $\rho\alpha\rho^{-1}(x) = x$.
2. $x \in \rho \text{Fix}(\alpha)$ means $x = \rho y$ for some $y$ fixed by $\alpha$.

We want so show $1 \implies 2$. Now suppose $x \in \text{Fix}(\rho\alpha\rho^{-1})$, so that $\rho\alpha\rho^{-1}(x) = x$. Then $\alpha \rho^{-1}(x) = \rho^{-1}(x)$. So what's $\alpha$ doing? It's fixing an element! how does the element that $\alpha$ is fixing relate to the element $x$?