I'm sorry if this is such silly question, but I can't seem to prove how the presentation of $D_{2n}$ keeps $r \neq s$ for $n = 2$. We have $$ \langle r,s \mid r^2 = s^2 = e, sr^{-1} = rs \rangle $$
If they're equal, then $r = r^{-1} = s$, so $$ rs = sr^{-1} = e. $$ I can't find a contradiction; where am I going wrong? If this is the Klein four group, I must not be getting something simple since it looks like can also generate $\langle r \rangle$.
I guess, the point is that there's a group, the Klein four group, in which $r\neq s$, and that group satisfies all the relations.
In a real sense, you quotient $F_2$, the free group on $2$ generators, by the smallest normal subgroup containing the relators.
That means you take the largest group possible. (And it's easy to show that the group has order at most four, using the anti-commutativity relation: $aba=b^{-1}$.)
So the group has order four. And is not cyclic (no element of order four).