If $a$ commutes with $b$, does $a^{-1}$ commute with $b$?
Solution 1:
$$\begin{align*} ab=ba &\Leftrightarrow aba^{-1}=baa^{-1}\\ &\Leftrightarrow aba^{-1}=b\\ &\Leftrightarrow a^{-1}aba^{-1}=a^{-1}b\\ &\Leftrightarrow ba^{-1}=a^{-1}b \end{align*}$$
Solution 2:
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: $$ba^{-1}=(a^{-1}ba)a^{-1}=a^{-1}b(aa^{-1})=a^{-1}be_G=a^{-1}b$$
Solution 3:
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 $a\in N_G(U)\implies a^{-1}N_G(U)$.
Solution 4:
@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.