How can we distinguish elements of $D_n$ that include reflections versus those that don't?

36 Views Asked by At

For $n \geq 3$, the dihedral group $D_n =\langle r, s \rangle$, where $r ^n = s^2 = e$.

Within this group, we can distinguish two types of elements:

  1. Those of the form $r^i$, where $i$ is any integer
  2. Those of the form $sr^i$, where $i$ is any integer

The latter includes not only $sr^i$ but any element with an odd number of $s$ in it. Examples are $r^is, sr^isr^js,$ etc.

What is a good way to distinguish this second class?

What is the best way to define it? The definition above is cumbersome. Can it be defined directly?

And: Is there a name for it?

1

There are 1 best solutions below

0
On BEST ANSWER

For odd $n$, the elements of order $2$ are precisely the elements of the form $sr^i$, $0\leq i\lt n$.

For even $n$, the non-central elements of order $2$ are precisely those of the form $sr^i$, $0\leq i\lt n$. The only element of order $2$ that is central is $r^{n/2}$.

Indeed, every element of the form $sr^i$ is of order $2$: $(sr^i)^2 = sr^isr^i = ssr^{-i}r^i = s^2=1$. If $r^i$ has order $2$, then $2i=n$. Thus, for odd $n$ these are the only elements of order $2$.

For even $n$, the exception is $r^{n/2}$; but this element is central: it commutes with $r$, and $sr^{n/2} = r^{-n/2}s = r^{n/2}s$, so it also commutes with $s$. And elements of the form $sr^i$ are not central: $(sr^i)r = sr^{i+1}$, and $r(sr^i) = sr^{-1}r^i = sr^{i-1}$. If these were equal, then $r^{i+1}=r^{i-1}$, so $r=r^{-1}$; but this requires $n=2$.