I mentioned in my last post that we had discussed exercise 42 from chapter 2 in our last meeting and that Dr. Prudhom gave me a hint on a tactic I could take to solve it. I tried using that method, but I found it to be actually much more difficult than I had anticipated—maybe I went about it in the wrong way, but the algebra was incredibly messy and felt much more convoluted than necessary. So I came back to the original statement of the exercise and tried to approach it from a group-theoretic perspective.

The first thing I did was to look up some examples of dihedral groups online, using a very handy website I found called GroupProps, which is basically “Wikipedia for groups.” You can find pages on a huge variety of groups that describe their properties and definitions. I looked up the dihedral groups, and specifically, the dihedral groups with orders that are multiples of 4 (i.e. the symmetry groups of regular polygons with an even number of vertices). I arrived at this restriction because I noted that the exercise implies that a rotation by 180° is an element of the group, and this is only the case in these specific dihedral groups. After all, if you rotate a triangle by 180° about its center, you don’t end up with a symmetry.

I noticed that in their presentations of these groups, GroupProps lists each reflection as a product of some “base” reflection with some number of rotations. That is, there are two basic, essential elements to the group, from which every other element can be constructed. (These are called a generating set of the group.) This led me to believe, although I have not been able to prove it, that any reflection in the dihedral group of order 2*k *can be written as the product of a single reflection *x* and some number of rotations by 360°*/k* degrees. (Call one rotation of 360/*k*° *a*.)

This observation lets me restate the problem (working in the dihedral group of order 2*k*) as follows:

I can then manipulate this expression into this form without changing the value of either side of the equation:

Note that I multiplied both sides by the inverse of x^2. But x^2 is the identity, so its inverse is also the identity, so I actually didn’t change the value of either side of the equation! This is important. (Just as a note, this works specifically because *x* is its own inverse, so I could have just replaced *x* with *x*^-1 without further justification. But this makes it even clearer that this step is justified.)

Once I have it in this form, I can use a second observation I gleaned from the GroupProps website, specifically, that

(Again, I don’t know how to prove it, but it makes intuitive sense if you play around with some concrete examples to see it in action.)

Edit (Jan. 31): I did find a way to prove it, and it’s actually really easy and I don’t quite see how I missed it. There’s absolutely no trick to it, all you have to do is write out the multiplication:

The last equality holds because the *x*s cancel with the *x*^-1 s that are right next to them.

That fact lets me rewrite the equation as follows, again, without changing the value of either side:

Now I just have to solve the modular congruence *n-m=m-n* mod 2*k*, which I’m comfortable doing thanks to my GOA Number Theory course last semester. I get that either *n=m *mod 2*k*, or *n-m=m-n=**k* mod 2*k*. But the first possibility is actually not a possibility, because we assumed that the two rotations were distinct, and this would imply that they are the same rotation. So we have that *n-m=m-n=**k* mod 2*k*.

Recall that I never changed the value of either side of the equation through all of my manipulations. Considering just the left hand side of all of the equations, this means that I have

My notation was different from the book’s notation, but this equality is exactly what I was asked to prove. I included the book’s notation under my equation for clarity.

I’m not sure that I took the most direct method, and I know that there are a lot of “holes” in my reasoning, insofar as there are facts that I used that I haven’t seen proven. But I think that this method does work regardless.