0
$\begingroup$

I'd like to request a hint

So far, I've enumerated all 16 symmetries. I know that $D_8$ has order $16$, so I've got them all. I know that I could in theory just check them all, which would be tedious but doable. I'm sure there is a trick to this though. Unfortunately, I do not have any other ideas. I know that in general, a rotation and a reflection do not commute, but I'm not sure if I can extend that to every rotation and reflection.

Obviously I have that the identity commutes.

If someone could give a hint that would be much appreciated.

$\endgroup$
1
$\begingroup$

Note that any diehedral group $D_n$, can be described as being generated by a generator $a$ of order $n$ and a generator $b$ of order $2$, bound by the relation $b^{-1}ab=a^{-1}$. Let $a^i$ be an element of the center of $D_n$ (i.e. an element that commutes with all other elements). It certainly already commutes with all powers of $a$ so we have to inspect when it commutes with $b$. This happens when $b^{-1}a^ib = a^{i}$ or $a^{-i} = a^{i}$, or $a^{2i}=1$. So $n$ has to be even and the commuting rotation is given by $a^{\frac{n}{2}}$, which correponds to a $180°$ turn.

$\endgroup$
  • $\begingroup$ This was very helpful. Thank you! How would I go about proving that reflections can never commute? $\endgroup$ – Alex Oct 6 '16 at 15:58
  • $\begingroup$ Every reflection has the form $s = a^ib$. So the condition for s to commute with every rotation $a^j$ is $s^{-1}a^js = a^j$, so we must have $b^{-1}a^{-i}a^ja^ib = a^j$, or $b^{-1}a^jb = a^j$ or $a^{-j} = a ^j$. This is clearly not true for all possible $j$. PS: I would much appreciate it if you mark my answer as ``accepted''. $\endgroup$ – Marc Bogaerts Oct 6 '16 at 17:54
1
$\begingroup$

Draw some pictures and try to understand why exactly the rotations and reflections do not commute (do it for a general regular polygon, boring otherwise). This will help you to understand what kind of rotation you should pick up as a candidate for the central (commuting with everyone) element.

$\endgroup$
  • $\begingroup$ Ok. After playing around with a cut out octagon for a while, I'm pretty sure that rotating $180^{\circ}$ works. I still really have no clue how to prove that in any way other than checking it for every other element. Not to mention that I still can't even convince myself! I don't think that there are any reflections that will work, but again I'm not able to prove this. $\endgroup$ – Alex Oct 5 '16 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.