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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

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    What did you try? – 2017-02-15
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    @PedroTamaroff I tried using the definition of the fixed set on permutations. So, ραρ-1(x)=x for all x. I tried manipulating this a few different ways to arise at the conclusion that ρα(x)=ρ(x) but could not do it. – 2017-02-15
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    Since you are trying to show equality of sets, showing containment both ways, might be the best way to go about it. – 2017-02-15
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    so show that x is in both by first assuming x is in fix(ραρ-1) and then again by assuming x is in ρ(fix(α))? – 2017-02-15
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    I understand how to show both ways, but I keep getting stuck with the algebraic manipulation... – 2017-02-15

1 Answers 1

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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 $x$?

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    oh I see what you are doing here. Well, since ρ−1(x) is fixed by α, y=ρ−1(x). So, x=ρy=ρρ−1(x)=x?!? How would I conclude this way of the argument more formally? – 2017-02-15
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    Exactly! By showing that $x$ has the form $\rho y$ for $y$ fixed by $\alpha$, you showed that $x$ is in the second set. – 2017-02-15
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    for 2 to 1, can you choose such a y? – 2017-02-15
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    For $2$ to $1$, you start by assuming $x \in$ second set, which means $x = \rho y $ for $y$ fixed by $\alpha$. What does it mean to show $x$ is in the first set? – 2017-02-15
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    to show x is in the first set.. we must show that ραρ-1(x)=x .. but how? – 2017-02-15
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    Yes, that's right. And we have $x = \rho y$. So what is $\rho \alpha \rho^{-1} (x)$ – 2017-02-15
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    ραρ-1(x)= ρρ-1(x)=x since α fixes ρ-1(x) – 2017-02-15
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    Exactly! that's what you wanted! – 2017-02-15
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    Oh right. So ραρ-1(x)= ραρ-1(ρy)=ραy=ρy=x ! – 2017-02-15
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    Yes, you've got it now. Keep working on formal manipulations ! It will come in time. And make sure you're always able to translate in between verbal and formal descriptions. – 2017-02-15
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    Thank you for all the help, I really appreciate it – 2017-02-15