Why must a group with presentation $\langle r,s \mid r^n=s^2=1,rs=sr^{-1}\rangle$ have exactly $2n$ elements? In Dummit and Foote's Abstract Algebra, they state on page 27 of chapter 1 that any group with the presentation $\langle r,s \mid r^n=s^2=1,rs=sr^{-1}\rangle$ must have exactly $2n$ elements. Why is this the case? If $G$ is the trivial group $\{1\}$, then it is surely the case that $1^n=1^2=1$ and $1\cdot 1=1\cdot 1^{-1}$, and $G$ is generated by $\{1\}$. It seems that I am misunderstanding how group presentations are defined, but I don't exactly know where my confusion lies.
 A: A group with presentation $\langle a_1,\dots,a_n\,|\, r_1,\dots,r_m\rangle$ is not just any group generated by elements $a_1,\dots,a_n$ which satisfy the relations $r_1,\dots,r_m$. It is the most general group satisfying those, in the sense that any group which satisfies those conditions is a quotient of that group. It is actually unique up to isomorphism.
You can either characterize it by this so-called universal property, or you can define it more prosaically as the quotient of the free group $F_n$ on the $n$ elements $a_1,\dots,a_n$ by the normalization of the subgroup generated by the relations $r_1,\dots,r_m$. (This construction gives a unique representative in the isomorphism class).
It is true that the trivial group satisfies any presentation (since it satisfies trivially any relation), and therefore it is (obviously) a quotient of the group actually given by the presentation.
A: A general element of the group has the form $r^is^j$ with $0\leq i\leq n-1$ and $s=0,1$, by the two relations. Thus the group has $2n$ elements.
