Intuition - $fr = r^{-1}f$ for Dihedral Groups - Carter p. 75 Name $r$ = clockwise 90 deg. rotation and $f$ = flip across the square's vertical axis = the brown $\color{brown}{f}$ in my picture underneath. Zev Chonoles's $f$ is different. Carter fleshes out why $frf = r^{-1} $ intuitively:



(1.) Can someone please unfold, like Carter, why $fr = r^{-1}f $? I see why for this case with $D_4$ but don't grasp the reason:

(2.) I tried algebra. Right-multiply: $frf\color{magenta}{f^{-1}} = r^{-1}\color{magenta}{f^{-1}} \iff fr = r^{-1}\color{magenta}{f^{-1}}  $.  What did I bungle?
 A: You've already said that you're using $r$ to denote clockwise rotation by $90^\circ$. Let's agree that the flip $f$ means "flip across its horizontal axis", so that
$$\fbox{$\begin{matrix}
1 & 2\\ 3 & 4
\end{matrix}$}\;\; \xrightarrow{f}\;\;\fbox{$\begin{matrix}
3 & 4\\ 1 & 2
\end{matrix}$}$$
(Note that the symbols are there only to denote which corner is which; they are not a part of the figure itself, and so are not getting flipped or rotated.)
As others have mentioned, $f^2$ is the identity (i.e., the "do nothing" operation), so that $f=f^{-1}$.

Observe that $frf$ does this:
$$\fbox{$\begin{matrix}
1 & 2\\ 3 & 4
\end{matrix}$}\;\; \xrightarrow{f}\;\;\fbox{$\begin{matrix}
3 & 4\\ 1 & 2
\end{matrix}$}\;\; \xrightarrow{r}\;\;\fbox{$\begin{matrix}
1 & 3\\ 2 & 4
\end{matrix}$}\;\; \xrightarrow{f}\;\;\fbox{$\begin{matrix}
2 & 4\\ 1 & 3
\end{matrix}$}$$
and $r^{-1}$ (counter-clockwise rotation by $90^\circ$) does this:
$$\fbox{$\begin{matrix}
1 & 2\\ 3 & 4
\end{matrix}$}\;\; \xrightarrow{r^{-1}}\;\;\fbox{$\begin{matrix}
2 & 4\\ 1 & 3
\end{matrix}$}$$

Observe that $fr$ does this:
$$\fbox{$\begin{matrix}
1 & 2\\ 3 & 4
\end{matrix}$}\;\; \xrightarrow{r}\;\;\fbox{$\begin{matrix}
3 & 1\\ 4 & 2
\end{matrix}$}\;\; \xrightarrow{f}\;\;\fbox{$\begin{matrix}
4 & 2\\ 3 & 1
\end{matrix}$}$$
and $r^{-1}f$ does this:
$$\fbox{$\begin{matrix}
1 & 2\\ 3 & 4
\end{matrix}$}\;\; \xrightarrow{f}\;\;\fbox{$\begin{matrix}
3 & 4\\ 1 & 2
\end{matrix}$}\;\; \xrightarrow{r^{-1}}\;\;\fbox{$\begin{matrix}
4 & 2\\ 3 & 1
\end{matrix}$}$$
A: $f$ is a reflection, so $f=f^{-1}$.
I think is okay now
A: Intuitively, flipping twice reverts to the original element, i.e. in algebra, ff = Id (identity), or $f^{-1}=f$ so from
$$
frf=r^{-1}
$$
You get, by "f 'ing" (pardon my French sense of humour :-) again :
$$
frff=r^{-1}f
$$
Thus
$$
fr=r^{-1}f
$$
(Funny thing, thix LaTeX-ish syntax)
