I am solving the following question from Aluffi. I am stuck on showing that this is the actual presentation of $D_{2n}$.
Question:
Describe generators and relations for dihedral groups $D_{2n}$.
My solution:
Claim:
$D_{2n} = \ \langle r,s : r^n = s^2 = e, rs = sr^{-1} \rangle$
I am confused on how to show that this generates $D_{2n}$. The book gives the following hint:
To see that these relations really determine $D_{2n}$, use them to show that any product $x^{i_1}y^{i_2}x^{i_3}x^{i_4}\ldots$ equals $x^iy^j$ for some $i,j$. The thing I am confused about is that how do we show this do we have to proceed by induction on $i_m$. I haven't seen this before. Can someone provide me with worked up version for $D_{2n}$?
The part you are asking about doesn't require induction. The trick is that $rs=sr^{-1} $ is an "anti-commutativity" relation, and allows you to switch the order of $r $ and $s$ (at a price, so to speak).
To take a very simple example, $srs=s(rs)=s (sr^{\color{blue}{-1}})=s^2r^{-1}=r^{n-1}$.
It's pretty clear that this sort of swapping will enable us to gather all the $s $'s and $r $'s together, so that everything can be written in the form $s^ar^b $.
Incidentally this immediately gives a bound on the size of the group.