# Dihedral Group $D_4$

1. Oct 30, 2014

### Bashyboy

In my class, we have denoted the elements of the dihedral group $D_4$ as $\{R_0, R_{90}, R_{180}, R_{270}, F_{\nearrow}, F_{\nwarrow}, F_{\leftrightarrow}, F_{\updownarrow} \}$. Not surprising, I was rather bewildered when I searched the internet for information on this group and encountered other denotations of the elements of $D_4$.

The standard denotation appears to be using $r$'s and $s$'s. I figured that $r = R_{90}$, $r^2 = R_{180}$, and $R_{270} = r^3$. However, I was unsure of the correspondence between the other elements. Could someone possibly help me with this?

2. Oct 31, 2014

### jbunniii

Assuming $R_{90}$ denotes counterclockwise rotation by $90$ degrees, and we use the convention that $ab$ means "do $b$ first, then $a$", you can check that
$R_{90} F_{\updownarrow} = F_{\nwarrow}$
$R_{180} F_{\updownarrow} = F_{\leftrightarrow}$
$R_{270} F_{\updownarrow} = F_{\nearrow}$
so if we put $r = R_{90}$ and $f = F_{\nwarrow}$ then the group is generated by $r$ and $f$. Moreover, $r$ has order $4$ and $f$ has order $2$, and you can verify that $f r f = r^{-1}$; indeed, the relations $r^4 = f^2 = e$ and $frf = r^{-1}$ suffice to define the group.

3. Nov 2, 2014

### Bashyboy

Okay, but what is the correspondence between $s$ and the flips (or reflections)?

4. Nov 2, 2014

### jbunniii

You can take $s$ to be any one of the four flips/reflections, because all four of them satisfy $s^2 = e$ and $srs = r^{-1}$. Try it with a square sheet of paper to verify this.

To verify algebraically that it doesn't matter which flip we assign to $s$, first choose $s$ to be one of the flips and verify geometrically that it satisfies $s^2 = e$ and $srs = r^{-1}$. Then note that the other three flips are $rs$, $r^2s$, and $r^3 s$. Then if we set $f_k = r^k s$ for $k=0,1,2,3$, we compute
$$f_k^2 = f_k f_k = r^k s r^k s = r^k (s r^k s) = r^k r^{-k} = e$$
where the third inequality holds by repeatedly applying $s r s = r^{-1}$. So $f_k$ is its own inverse. Then
$$f_k r f_k = (r^k s) r (r^k s) = r^k (s r^{k+1} s) = r^k (r^{-(k+1)}) = r^{-1}$$
We have therefore shown that $f_k$ satisfies $f_k^2 = e$ and $f_k r f_k = r^{-1}$ for $k=0,1,2,3$, so any of the $f_k$ obeys the defining properties of $s$.

Last edited: Nov 2, 2014