If $a$ commutes with $b$, does $a^{-1}$ commute with $b$? $a,b$ are group elements. If $a$ commutes with $b$, does $a^{-1}$ commute with $b$?
And how can I prove this?
 A: @parsec Now you have seen all the answers you probably can prove that any power $a^n$ commutes with $b$, in other words, the group generated by your $a$ and $b$, $<a,b>$,  is abelian.
A: $$\begin{align*}
ab=ba
&\Leftrightarrow aba^{-1}=baa^{-1}\\
&\Leftrightarrow aba^{-1}=b\\
&\Leftrightarrow a^{-1}aba^{-1}=a^{-1}b\\
&\Leftrightarrow ba^{-1}=a^{-1}b
\end{align*}$$
A: Note that if $$ab=ba$$ then multiply previous relation by $a^{-1}$ from the left to get: $$b=e_Gb=(a^{-1}a)b=a^{-1}(ab)=a^{-1}(ba)=a^{-1}ba$$ Now , again multiply previous relation by $a^{-1}$ from the right to get: $$ba^{-1}=(a^{-1}ba)a^{-1}=a^{-1}b(aa^{-1})=a^{-1}be_G=a^{-1}b$$
A: More generally, for subsets $U\subseteq G$, the normalizer of $U$ is defined as
$$ N_G(U):=\{\,g\in G\mid gU=Ug\,\}$$
and is a subgroup of $G$ because 


*

*$eU =U=Ue$ trivially

*$gU=Ug $ and $hU=Uh$ implies $(gh)U=g(hU)=(gU)h=(Ug)h=U(gh)$

*$gU=Ug$ implies $g^{-1}U=g^{-1}Ugg^{-1}=g^{-1}gUg^{-1}=Ug^{-1}$


that is $N_G(U)$ is nonempty and closed under multiplication and taking inverses.
You are concerned with the special case $U=\{b\}$ and the claim $a\in N_G(U)\implies a^{-1}N_G(U)$.
