Intereting Posts

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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:

- On the centres of the dihedral groups
- Prove that the dihedral group $D_4$ can not be written as a direct product of two groups
- Prove two reflections of lines through the origin generate a dihedral group.
- Finding Sylow 2-subgroups of the dihedral group $D_n$
- Algebra - Infinite Dihedral Group
- How to describe all normal subgroups of the dihedral group Dn?

~~(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?

- Is there a geometric realization of Quaternion group?
- The product of finitely many cyclic groups is cyclic
- Is every finite group of isometries a subgroup of a finite reflection group?
- Square free finite abelian group is cyclic
- Computing the order of elements in Dihedral Groups
- Transitive subgroup of symmetric group $S_n$ containing an $(n-1)$-cycle and a transposition
- Quotient groups of a finite symmetric group
- $|G|=12$ and it is isomorphic to $A_4$?
- Finite groups with exactly one maximal subgroup
- $A_4$ has no subgroup of order $6$?

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}$}$$

$f$ is a reflection, so $f=f^{-1}$.

I think is okay now

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 ðŸ™‚ again :

$$

frff=r^{-1}f

$$

Thus

$$

fr=r^{-1}f

$$

(Funny thing, thix LaTeX-ish syntax)

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