User:IssaRice/Generators and relations of dihedral groups

problem II.2.5 (p. 57) in Aluffi's Algebra: Chapter 0.

As suggested by the hint, take $x$ to be a reflection through the center and some vertex, and $y$ to be a counter-clockwise rotation of $2\pi /n$. Doing the same reflection puts the n-gon back in place, so $x^{2}=e$, and rotating $n$ times also puts the n-gon back in place, so $y^{n}=e$. Finally, doing the suggested hand trick suggests $xyxy=(xy)^{2}=e$ (here we start applying the operations on the left, so this is the opposite of the usual functional application notation). But written in this form, it's not so obvious that this all we need to determine the group. So it would be nice if we could convert $(xy)^{2}=e$ to an expression involving $yx$ on one side (this allows us to flip x and y whenever they appear in the wrong order). Playing around, we get

$${\begin{aligned}x(xyxy)y^{n-1}&=x(e)y^{n-1}\\(x^{2})(yx)y^{1+n-1}&=xy^{n-1}\\yx&=xy^{n-1}\end{aligned}}$$

Now given $x^{i_{1}}y^{i_{2}}x^{i_{3}}y^{i_{4}}$, we can simplify the powers on x using $x^{2}=e$ and the powers on y using $y^{n}=e$, so we can assume we've done that, and are left with a power of 0 or 1 on each x, and a power that's some integers in $[0,n)$ on the y. splitting into cases based on $i_{3}$:

• $i_{3}=0$ -- we have $x^{i_{1}}y^{i_{2}}y^{i_{4}}$, and this simplifies to $x^{i_{1}}y^{i_{2}+i_{4}}$. now just use $y^{n}=e$ to simplify.
• $i_{3}\neq 0$ -- we have $x^{i_{1}}y^{i_{2}}xy^{i_{4}}$. focus on $y^{i_{2}}x$. by an $i_{2}$-times application of the relation $yx=xy^{n-1}$, we get $xy^{i_{2}(n-1)}$. for example, if $i_{2}=3$, we would have $y^{3}x=y^{2}(yx)=y^{2}(xy^{n-1})=y(yx)y^{n-1}=y(xy^{n-1})y^{n-1}=(yx)y^{2(n-1)}=(xy^{n-1})y^{2(n-1)}=xy^{3(n-1)}$. so in the end we get $x^{i_{1}}y^{i_{2}}xy^{i_{4}}=x^{i_{1}}xy^{i_{2}(n-1)}y^{i_{4}}$. now simplify using the rules involving just x and y.

For a product involving more than four powers of x and y, just work on it four powers at a time. do the first four, which turns into just two powers by the algorithm above. then bring in the next two, and simplify using the same algorithm above. then the next two, and so on. if the final power has just x, we can always pretend y has exponent 0.

Why is it obvious that x and y generate the group? the vertices of the n-gon have to come in order (otherwise you are disfiguring the n-gon somehow), so there is a clockwise version and a counter-clockwise version to order the vertices, and a reflection can take us between them. then for each case, all we can do is decide where the first vertex lands -- once we do that, everything else has to count up (in either clockwise or counter-clockwise direction). there are n choices for the first vertex. so there are 2n total symmetries.