Presentation of Dihedral Group

Question

Consider the standard presentation of $D_{2n}$:

$\langle r, s : r^n = s^2 =1, rs = sr^{-1}\rangle$.

I have seen the latter relation given as $sr = r^{-1}s$ a few times. Is this correct, as well?

Answer 1

It is correct

$$rs = sr^{-1} \Leftrightarrow rsr = s \Leftrightarrow sr = r^{-1}s$$

Answer 2

I would believe it simply because the group defined "doing everything backwards" is isomorphic to the group you get "doing everything normally."

In particular, given a group $(G, *)$, define its opposite group $(G^{\rm op}, *^{\rm op})$ by $g *^{\rm op} h = h * g$. It's a fun exercise to show that the map $G \to G^{\rm op}$ defined by $g \mapsto g^{-1}$ is an isomorphism.