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I'm trying to show that if $Y \subseteq $ a group $G$ and $z \in G$, then $C_{G}(z^{-1}Yz)= z^{-1}C_{G}(Y)z$, where $C_{G}(Y)$ is the centralizer of $Y$ (i.e., $C_{G}(Y) = \{x \in G | xy = yx, \forall y \in Y \}$).

I understand that since this is a "set equals", I need to show that $C_{G}(z^{-1}Yz)\subset z^{-1}C_{G}(Y) z$ and $z^{-1}C_{G}(Y) z \subset C_{G} (z^{-1}Yz)$.

In the $z^{-1}C_{G}(Y)z \subset C_{G}(z^{-1}Yz)$ direction, let $w \in z^{-1} C_{G}(Y) z$. Then, $z^{-1}(wy)z = z^{-1}(yw)z$, $\forall y \in Y$. Then, we want to show that $w(z^{-1}Yz) = (z^{-1}Yz)w$, but I am very confused with how to show this. Multiplying both sides of $z^{-1}(wy)z = z^{-1}(yw)z$ by things isn't getting me anywhere, and I am a loss at what to try next. Sometimes these proofs involve clever little tricks I wouldn't have thought of, and so I am asking whether I am missing something here, and if so (and even if not!) how I should proceed.

In the $C_{G}(z^{-1}Yz) \subset z^{-1}C_{G}(Y)z$ direction, if we let $u \in C_{G}(z^{-1}Yz)$, then $uz^{-1}Yz =z^{-1}Yzu $, and we want to show that this implies that $u \in z^{-1}C_{G}(Y)z$, or $z^{-1}(uy)z = z^{-1}(yu)z$ for all $y \in Y$. Again, same problem as before.

I've seen proofs showing this for the Normalizer, but I haven't understood the inclusions in the $N_{G}(z^{-1}Yz) \subset z^{-1}N_{G}(Y)z$ direction. Anyway, even if I could understand them, in order to use that to help me prove this, since the centralizer is contained in the normalizer, I'd have $C_{G}(z^{-1}Yz) \subset N_{G} (z^{-1}Yz)$ and $z^{-1}C_{G}(Y) z \subset z^{-1}N_{G}(Y)z$, but then how would I determine anything about the relationship between $N_{G} (z^{-1}Yz)$ and $z^{-1}C_{G}(Y)z$? (And the relationships in the other direction as well?)

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    More generally, if $G$ acts on a set $X$, then ${\rm Stab}(g\cdot x)=g{\rm Stab}(x)g^{-1}$ for all $g\in G,x\in X$. In this case, we have $G$ acting on itself by conjugation. – 2017-01-02
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    @arctictern I'm not familiar with $Stab(gx)$. – 2017-01-02
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    @JessyCat It's the [stabilizer subgroup](https://en.wikipedia.org/wiki/Group_action#Orbits_and_stabilizers). – 2017-01-02
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    @SpamIAm oh I got that, but for us that's in the next chapter, so I can't use it here. It's very edifying to see, but right now I'm more concerned about the task I have to complete right now. – 2017-01-02

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Suppose $w\in zC_G(Y)z^{-1}$. Therefore $w=zxz^{-1}$ for some $x\in C_G(Y)$. We want to verify that $w$ is also an element of $C_G(zYz^{-1})$, so let's pick an arbitrary $zyz^{-1}\in zYz^{-1}$ (where $y\in Y$) and,

$$ \begin{array}{ll} w(zyz^{-1}) & =(zxz^{-1})(zyz^{-1}) \\ & =z(xy)z^{-1} \\ & =z(yx)z^{-1} \\ & =(zyz^{-1})(zxz^{-1}) \\ & =(zyz^{-1})w. \end{array} $$

Since $w$ commutes with any element $zyz^{-1}$ of $zYz^{-1}$, we have that all $w\in zC_G(Y)z^{-1}$ are members of $C_G(zYz^{-1})$, or in other words we have the containment $zC_G(Y)z^{-1}\subseteq C_G(zYz^{-1})$.

To prove the other other direction, rewrite the containment we just did, replacing $Y$ with $z^{-1}Yz$ and then conjugating by $z$.

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    The other direction is proved by replacing $z$ with $z^{-1}$ and $Y$ with $zYz^{-1}$ in the last containment, then conjugating by z. – 2017-12-17
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$$h\in C_G(Y) \leftrightarrow [h,Y]=1 \leftrightarrow z^{-1}[h,Y]z=1 \leftrightarrow [z^{-1}hz, z^{-1}Yz]=1\leftrightarrow z^{-1}hz\in C_G(z^{-1}Yz)$$

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$x \in C(y) \leftrightarrow x^{-1} y x = y.$ let $z = u y u^{-1},$ then $u y u^{-1} \in C(z).$ The other direction is by definition.

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    $x^{-1}yx=y$ implies that $x \in $ the normalizer of $y$, doesn't it? – 2017-01-02
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    Elements don't have normalizers, they only have centralizers, which are defined exactly as I said. – 2017-01-02
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    also, how does $z=uyu^{-1}$ imply that $uyu^{-1} \in C(z)$? There are so many little things going on behind the scenes that I am obviously missing... – 2017-01-02
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    Igor, I think you mean $uxu^{-1}\in C(z)$. – 2017-01-03
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    @Jessy Suppose $x\in C(y)$. Have you tried checking of $uxu^{-1}$ commutes with $uyu^{-1}$? That is the condition for $uxu^{-1}$ being a member of $C_G(uyu^{-1})$, so that's the thing you have to check. Whenever you do these problems, repeat these two mantras infinitely: "What is it asking me to do now?" and "What does that mean?" – 2017-01-03