Name the corners of a square as 1,2,3,4 in clockwise order.

As you know: The group of all rigid motions of the square back to itself, called $D_4$, has eight elements, written in cycle form as: $$\begin{align*} \mathrm{id},& &\qquad a&=(1234),\\ b&=(13)(24), &c&=(1432),\\ d&=(12)(34), &e&=(14)(23),\\ f&=(13), &g&=(24). \end{align*}$$

The two subgroups $x_1= \langle d, e\rangle$ and $x_2=\langle f,g\rangle$ are each "abstractly" isomorphic to the Klein 4-group, and share the central element $b$.

But $x_1$ and $x_2$ are not conjugate as subgroups of $D_4$ (or as subgroups of $S_4$, since they have different cycle structure, and half of $x_2$ is odd while all of $x_1$ is even).

$D_4$s regular representation in $A_8$ can be written as $$\begin{align*} \mathrm{id}, &&\quad A&=(1234)(5768);\\ B&=(13)(24)(56)(78);& C&=(1432)(5867);\\ D&=(15)(28)(36)(47);& E&=(16)(27)(35)(48);\\ F&=(17)(25)(38)(46);& G&=(18)(26)(37)(45). \end{align*}$$ The image of $x_1$ is $\langle D,E\rangle$ and that of $x_2$ is $\langle F,G\rangle$.

But, $\langle D,E\rangle$ is automorphic to $\langle F,G\rangle$ under the conjugation in $S_8$ by $(5768)$.

My question: Is there a way to draw $D_4$ as motions of a geometric figure that would allow me to see the automorphism?



It may not be what you want, but take a cube, and number the vertices on top $1,2,3,4$, and the vertices directly below them $7,6,8,5$, like so:

                  /|        /|
                 / |       / |
                3_________4  |
                |  |______|__|7
                |  /6     |  /
                | /       | /
                |/        |/

You can now consider the rigid motions of the cube that respect "sides" and "top-and-bottom". That is, the top and bottom face must be mapped to the top and bottom face, the four sides to the four sides.

$\mathrm{id}$, $A$, $B$, and $C$ are rotations that map the top face to itself; $D$, $E$, $F$, and $G$ are the rotations that exchange top and bottom faces. The automorphism corresponds to renumbering the vertices in the bottom.


You can draw two squares inside a regular octagon. If the vertices of the octagon are (in clockwise order) $1,1',2,2',3,3',4,4'$, then the two squares are formed by the unprimed and primed vertices respectively. If you rotate the whole picture by $2\pi/8$, then the two squares are interchanged.

The two squares give rise to the same dihedral group $D_4$ (when viewed as motions of the plane): four rotations (one trivial) and four reflections. The $x_1$ of one square looks like the $x_2$ of the other square. In the image below the two squares are colored red (unprimed) and blue (primed) rescpetively. Imagine that $1'$ is the top corner of the blue square, so $3'$ is at the bottom. Orient it so that $1,2,3,4$ are the NW, NE, SE, SW corners respectively. The reflection w.r.t. the vertical axis gives $d=(12)(34)$ on the red square, and $g'=(2'4')$ on the blue square. Similarly other elements of $x_1$ on the red square become elements of $x_2$ on the blue square and vice versa.

A regular octagon with two inscribed squares

Another way of looking at it is, if $r$ is a rotation by the angle $2\pi/8=\pi/4$, you get $x_1$ from $x_2$ (of the same square) by conjugating it with $r$. In other words, the two subgroups become conjugate, if you view both of them as subgroups of $D_8$.

  • 1
    $\begingroup$ Or if $b$ is the reflection wrt the $y$-axis, and $r$ the clockwise rotation by 45 degrees, then $rbr^{-1}$ is reflection wrt to the line $y=x$, which is either $f$ (or $g$, if I made a mistake while labelling). Both $b$ and $rbr^{-1}$ are in the group $D_4$, but $r$ is not. To make $r$ into a symmetry, we need the octagon. $\endgroup$ – Jyrki Lahtonen Mar 30 '12 at 19:54

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