Here's a problem form I. Martin Isaacs Finite Group Theory (exercise 1A.10.(a)):
Let $H$ be a subgroup of $G$. Show that $|\textbf{N}_G(H):H|$ is the number of right cosets of $H$ in $G$ invariant under right multiplication by $H$.
Cihan Bahran has uploaded the solutions to the exercises of this book, and he solves this problem as follows:
Let $\Omega$ be the set of right cosets of $H$. $H$ acts on $\Omega$ by right multiplication. Denoting the set of fixed points by $\Omega_0$, we have: \begin{align*} \Omega_0&=\{Ha:Hah=Ha\text{ for all }h\in H\}\\ &=\{Ha:aha^{-1}\in H\text{ for all }h\in H\}\\ &=\{Ha:aHa^{-1}\subseteq H\}\\ &\color{red}{=}\{Ha:H^a=H\}\\ &=\{Ha:a\in \textbf{N}_G(H)\}\\ &=|\textbf{N}_G(H):H|. \end{align*}
However, I don't understand why equality in red holds. Because $aHa^{-1}\subseteq H$ does not, in general, imply that $H^a=H$. Can someone help me understand this step of the proof or, otherwise, provide another proof for this exercise? Thanks.
If $g\in N_G(H)$, then the right coset $Hg$ is fixed by $H$. The reverse implication is not necessarily true when $H$ is infinite.
Let $G=\{ax+b\mid a,b\in\mathbb{R}\}$ be maps on the real line, and $H=\{x+n\mid n\in\mathbb{Z}\}$. Then for $g=mx$ with $m$ an integer, $Hg$ is fixed by $H$. But for most $m$ , $g\not\in N_G(H)$.