Cayley table for $D_4/\langle r^2\rangle$.

Question

Write out the Cayley table for $D_4/\langle r^2\rangle$.

I know that $D_4$ is the dihedral group with 4 sides (square) and $\langle r^2\rangle$ is the algebra generated by $r^2$ (a.k.a. taking the powers of $r^2$) but I am unsure how this Cayley table is generated. My assumption is that since $r^2 = \{\mathrm{id}, r^2\}$, the Cayley table will just be filled with id and $r^2$'s? I would appreciate any feedback.

Answer 1

First of all, $\langle r^2\rangle$ is the subgroup generated by $r^2$ ( not the algebra.) You have to make sure that you understand what is $G=D_4/\langle r^2\rangle$.

This is a quotient group. It means that elements of $G$ are equivalence classes of elements of $D_4$ for the equivalence relation $x\sim y\iff x^{-1}y\in\langle r^2\rangle$. In this particular case, it means that $x\sim y\iff x=y$ or $x=yr^2$, as you noticed.

If $x\in D_4$, let us denote by $\overline{x}$ its equivalence class in $G$. Then, by properties of equivalence classes $\overline{x}=\overline{y}\iff x\sim y$.

Thus, in the quotient group, $r^2$ will "disappear" since $\overline{r^2}=\overline{Id}$. So you certainly won't have any $r^2$ in your Cayley table.

The first thing to do is to find the elements of $D_4$. You will have to use the fact that the elements are $r^ms^n, m=0,1,2,3, s=0,1$, where $s$ is an orthogonal reflexion wrt to the $x$-axis, for example, and $r$ is a rotation of angle $\pi/2$ (so $r$ has order $4,$ $s$ has order $2$ and $srs^{-1}=r^{-1}$).

You shoud find $4$ elements in $G$, since $D_4$ has order $8$ and $\langle r^2\rangle $ has order $2$ (which ones? Beware that you need to ensure that your elements are pairwise distinct!)

Then you will have to multiply them, and to write the product in such a way you find again one of your $4$ elements.

For example, imagine you want to multiply $\overline{r}$ and $\overline{r}$. We have $\overline{r} \ \overline{r}=\overline{r^2}=\overline{Id}$.

At then end, you should find a Cayley table which is similar to one group of order $4$ which is more familiar to you...

Answer 2

You will get the Klein four group, $\Bbb Z_2\times\Bbb Z_2$. This follows from the fact that $\langle r^2\rangle $ is the center of $D_4$, and if a group modulo its center is cyclic, then the group is abelian.

See here for the Cayley table. You can identify $s\leftrightarrow a, rs\leftrightarrow b$ and $r^3s\leftrightarrow c$, for instance.