If G is a group and $x,a \in G$, prove that $C_G(x^{-1}ax)=x^{-1}C_G(a)x$ If G is a group and $x,a \in G$, prove that $C_G(x^{-1}ax)=x^{-1}C_G(a)x$ where $C_G$ is the centralizer.
I believe we have to do this by showing that $C_G(x^{-1}ax) \subset x^{-1}C_G(a)x$ and vice versa. 
So, let $g \in C_G(x^{-1}ax)$,  
then $$gx^{-1}axg^{-1} = x^{-1}ax$$
$$gx^{-1}ax = x^{-1}axg$$
$$x^{-1}ax = g^{-1}x^{-1}axg$$
$$a = xg^{-1}x^{-1}axgx^{-1}$$
So $$xgx^{-1} \in C_G(a)$$
And, $$x^{-1}C_G(a)x = g$$
The other inclusion is just reverse of the argument. 
 A: Your argument is correct, you can prove both the inclusions by picking an arbitrary element of the LHS and showing that it's in the RHS and vice versa. It basically boils down to undoing the conjugation by $x$ or $x^{-1}$ . 
Here's another way to structure a proof using more or less the same trick.


*

*I'm using $\overline{g}$ to mean $g^{-1}$ . I think it reads better if you have lots of group elements being multiplied together.

*I'm using $\cdot$ to explicitly represent the result of multiplying a set of group elements by a single group element (elementwise product), e.g. $ g \cdot S = \{ gs \mathop| s \in S \} $ and $ S \cdot g = \{sg \mathop| s \in S\}$ . The elementwise product is normally written with juxtaposition: $gS$ and $Sg$ . $(\cdot)$ has lower precedence than juxtaposition.


We want to show that the negation of the original statement implies a contradiction.
$$ C_G(\overline{x} a x) \neq \overline{x} \cdot C_G(a) \cdot x \tag{NG} $$
$$ x \cdot C_G(\overline{x} a x) \cdot \overline{x} \neq x \overline{x} \cdot C_G(a) \cdot x \overline{x} \tag{101} $$
$$ x \cdot C_G(\overline{x} a x) \cdot \overline{x} \neq C_G(a) \tag{102} $$
replace $C_G$ on LHS with its definition.
$$ x \cdot \{ g \mathop| g \in G \land g\overline{x}ax = \overline{x}axg \} \cdot \overline{x} \neq C_G(a) \tag{103} $$
bring $x \dots \overline{x}$ inside the set builder notation.
$$ \{xg\overline{x} \mathop| g \in G \land g\overline{x}ax = \overline{x}axg \} \neq C_G(a) \tag{104} $$
Rewrite LHS using a new variable $h = xg\overline{x}$ .
$$ \{ h \in G \mathop| \overline{x}hx\overline{x}ax = \overline{x}ax\overline{x}hx \} \neq C_G(a) \tag{105} $$
$$ \{ h \in G \mathop| \overline{x}hax = \overline{x}ahx \} \neq C_G(a) \tag{106} $$
Replace the condition inside the LHS of (106) with the equivalent condition $ha = ah$ . This is allowed because conjugation is invertible.
$$ \{ h \in G \mathop| ha = ah \} \neq C_G(a) \tag{107} $$
Contradiction, (107) is the definition of the centralizer.
$$ \bot $$
