Simple proof of "$a$ and $a^{-1}$ have the same number of conjugates"

Question

I recently had to give a proof of this, I gave a correct proof but I feel that it was overly complicated, so the question here is "Find a simpler proof (one that belongs in "the book")".

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Question: "Given a group $G$ and $a\in G$, prove that $a$ and $a^{-1}$ have the same number of distinct conjugates."

My answer (just a sketch):

Take $Cl(a)$ (conjugacy class of a) and $Cl(a^{-1})$, show that $x\in Cl(a) \iff x^{-1}\in Cl(a^{-1})$ (call this Statement 1).

Form the function $g$ on a set $S\subset G$, $g(S)=\{s^{-1}|s\in S\}$. Show that this is an involution, and thus a bijection.

Show that $Cl(a^{-1})=g(Cl(a))$ (by proving both $Cl(a^{-1})\subset g(Cl(a))$ and $g(Cl(a))\subset Cl(a^{-1})$ , by using Statement 1).

By definition of "same cardinality" ($\equiv$ there exists a bijection between them), conclude $Cl(a^{-1})=g(Cl(a)) \implies |Cl(a^{-1})|=|Cl(a)|$.

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This took about 2.5 pages to fully explicate. Surely there must be a more concise proof for such a simple question?

Answer 1

Let $G$ act on itself by conjugation. The stabiliser of $a$ is $C(a)$, the centraliser of $a$, and $C(a) = C(a^{-1})$, so the number of conjugates of $a^{-1}$ is $(G:C(a^{-1})) = (G:C(a))$, which is the number of conjugates of $a$.

Answer 2

I believe you're overthinking. ;-)

The map $x\mapsto x^{-1}$ induces a map $\operatorname{Cl}(a)\to\operatorname{Cl}(a^{-1})$.

Indeed, if $x=gag^{-1}$ is a conjugate of $a$, then $x^{-1}=ga^{-1}g^{-1}$ is a conjugate of $a^{-1}$.

Call $f_a$ the map above. Then $f_{a^{-1}}\colon \operatorname{Cl}(a^{-1})\to\operatorname{Cl}(a)$ has the property that $f_{a^{-1}}\circ f_a$ is the identity on $\operatorname{Cl}(a)$. Similarly, $f_{a}\circ f_{a^{-1}}$ is the identity on $\operatorname{Cl}(a^{-1})$.

Note that this doesn't require $G$ to be finite.