TITLE: Let $a \in G$. Show that for any $g \in G$, $gC(a)g^{-1} = C(gag^{-1})$.
QUESTION [0 upvotes]: Let $a \in G$. Show that for any $g \in G$, $gC(a)g^{-1} = C(gag^{-1})$.

Note that $C(a) = \left \{g \in G : ag = ga \; or\; gag^{-1} = a\right \} $.
How do I begin such a problem? I thought about letting $ghg^{-1}\in gC(a)g^{-1}$. But I am not sure what to do afterwards.

REPLY [2 votes]: The standard "trick" for showing $A = B$ for sets $A, B$ is $A \subseteq B$ and $B \subseteq A$. We will employ that.
$\newcommand{\gi}{g^{-1}}
\newcommand{\bi}{b^{-1}}
\newcommand{\xi}{x^{-1}}
\newcommand{\seq}{\subseteq}$
Let $b = gx\gi \in gC(a)\gi$. Then $b(ga\gi)\bi = (bg)a(bg)^{-1} = gxa\xi\gi = ga\gi$ as $xa\xi = a$. So $b \in C(ga\gi)$. Since $b$ was general, $gC(a)\gi \subseteq C(ga\gi)$.
Now lets use another trick. We will not prove the other direction explicitly. Instead lets use the result we just proved.
$gC(a)\gi \seq C(ga\gi)$. Put $a = \gi c g$. So $gC(\gi cg)\gi \seq 
 C(c)$. So $C(\gi c g) \seq \gi C(c) g$, which is the reverse direction.
This is a standard trick that works in groups because of the bijective properties of the product.
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