Let $A=\{a\}$ with $a\in G$, a group. Prove $C_G(a) = C_G(a^{-1})$ Let $G$ be a group and let $ a \in G$. If $A = \{a \}$, prove $C_G(a) = C_G(a^{-1})$
Proof:
Since $C_G(a) \le G$, it is given that $a^{-1} \in C_G(a)$ Since $A=\{a\}$ it must be the case that $C_G(a) = C_G(a^{-1})$. since there are no other elements in A.
Okay --- so I am just not sure what exactly $A=\{a\}$ means. Is it the singleton set? Because when I read the text it says that $a^n \in C_G(a)$ for all $ n \in \mathbb{Z}$. I am not really certain where to start.
 A: For any subset $A \subset G$, we define $C_G(A)$ as
$$C_G(A):= \{x\in G| xa = ax \ \text{for all} \ a \in A \} $$
This is set of all elements in your group which commute with $A$. $C_G(a)$ is the same thing as $C_G(\{a\})$ in this case. So you want to show that $xa = ax$ if and only if $xa^{-1} = a^{-1}x$.
\begin{align*}
 xa & = ax  \\ xax^{-1} & = a \\ x a^{-1}x^{-1} & = a^{-1} \\ xa^{-1} & = a^{-1}x
\end{align*}
A: Claim. Let $G$ be a group and let $ a \in G$. If $A = \{a \}$, prove $C_G(a) = C_G(a^{-1})$.
Quick comment: Centralisers are typically defined for sets, but the notation $C_G(a)$ for $C_G(\{a\})$ is prevalent. The question probably means something like "prove $C_G(A) = C_G(\{a^{-1}\})$", which is where $A$ would be used, but then the author got confused. Or something. Anyway...
Proof of claim:
If $b\in C_G(a)$ then we have the following sequence of equivalent identities
\begin{align*}
ab&=ba\\
ab\cdot a^{-1}&=ba\cdot a^{-1}\\
aba^{-1}&=b\\
a^{-1}\cdot aba^{-1}&=a^{-1}\cdot b\\
ba^{-1}&=a^{-1}b
\end{align*}
Therefore, an element $b$ commutes with $a$ if and only if $b$ commutes with $a^{-1}$.
Since $A={a}$, $C_G(a)$ is precisely those elements of $G$ which commute with $a$, and so $C_G(a) = C_G(a^{-1})$ as required.
