Source: Abstract Algebra 3rd edition by Dummit & Foote.
At the end of Section 1.2, the author writes: "This kind of collapsing does not occur for the presentation of $D_{2n}$ because we showed by independent (geometric) means that there is a group of order 2n with generator r and s and satisfying the relations in (1). As a result, a group with only these relations must have order at least 2n."
By presentation, the author refers to $D_{2n}=\left\langle r, s\mid r^n=s^2=1, rs=sr^{-1}\right\rangle$; and by relation, the author refers to $r^n=s^2=1$ and $rs=sr^{-1}$.
My Question: I failed to follow this argument. For example, the dihedral group of order 8 indeed satisfies $r^8=s^2=1, rs=sr^{-1}$, which is the presentation for $D_{16}$; but clearly $|D_8|=8<|D_{16}|=16$.
What am I missing here? Any help would be greatly appreciated.
Best Answer
Given a presentation, there can many groups satisfying the presentation, but the presentation always define a unique group, which is in some sense the $\textbf{largest}$ one satisfying the presentation. So whenever a group of order $n$ satisfies a presentation, it only shows that the group defined by this presentation has order at least $n$, so it is a lower bound. To find an upper bound, one has to use a different kind of argument (and it's not always easy).