Determining a lower bound on the order of a group based on its presentation I am reading Abstract Algebra book by Dummit and Foote (3-rd edition). 
On pages 26-27 they define a dihedral group:
$D_{2n} = \langle r,s | r^n = s^2 = 1, rs = sr^{-1} \rangle$
The authors describe a group (a regular n-gon's rotations, where $r$ is $\frac{2\cdot\Pi}{n}$ rotation and $s$ is a flip) which satisfies this representation.
Then they claim that, based on the existance of such a group, any group with such a representation must have an order of at least 2n. Why is it so? Is there any theorem from which they derive the claim?

If you take $G_{2n}$ as the subgroup of orthogonal transformations (rotations, flip etc.), it is not hard to see that this group has indeed $2n$ elements, and that it is a homomorphic image of $D_{2n}$. (Given where the remark is in the book, this argument of homomorphic image is not entirely kosher.)
Thus the order of $D_{2n}$ needs to be at least equal to the order of $G_{2n}$.
