I'm going through Dummit and Footes Abstract Algebra book, and have a question on their characterization of group presentations.
They state that:
In general, if some group $G$ is generated by a subset $S$ and there is some collection of relations, say $R_1, R_2, \ldots , R_m$ (here each $R_i$ is an equation in the elements rom S $\bigcup$ {1} ) such that any relation among the elements of S can be deduced from these, we shall call these generators and relations a presentation of G and write $G = \langle S | R_1, R_2, R_3, … , R_m \rangle$.
He goes on to give the dihedral group presentation as:
$$ D_{2n} = \langle r,s | r^n = s^2 = 1, rs = sr^{-1} \rangle \space\space (1) $$
The authors had shown geometrically that $D_{2n}$ has order $2n$ but they went on to say that as a result any group with only the relations in (1) must have order at least $2n$. They also claim that any any group with the (1) presentation must also have order at most 2n ( the author states that using the relation $rs = sr^{-1}$ accomplishes this). Hence the order must equal $2n$.
Edit: To quote the others directly from page 27, chapter 1.2:
" This kind of collapsing does not occur for the presentation of D$_{2n}$ because we have showed by independent (geometric) means that there is a group of order 2n with generators r and s satisfying the relations in (1). As a result, a group with only these relations must have order at least 2n. On the other hand, it is easy to see (using the same sort of arguments for X$_{2n}$ above and the commutation relations rs = sr$^{-1}$) that any group defined by the generators and relations in (1) has order at most 2n. It follows that the group with presentation (1) has order exactly 2n and also that this group is indeed the group of symmetries of the regular n-gon."
Can someone please provide the details behind this argument for why the presentation guarantees the order to be equal to $2n$? Why do we have that if a group $G$ ONLY has the relations in (1) then we must have $|G| \ge 2n$? If such an argument works to show $|G| \ge 2n$, why doesn't that same argument work for showing $|G| \le 2n$?
Best Answer
Here's a little elaboration: Any element of the group $G$ given by the presentation above may be thought of as a word, that is, a finite sequence in $r$ and $s$ and their inverses.
Notice, the relation $rs = sr^{-1}$ allows us to progressively rearrange our words so that they are of the form $s^kr^{\ell}$. For example, $$\begin{align}rsr^2s^{-3}r &\leadsto sr^{-1}r^2s^{-3}r^4 = srs^{-3}r^4 \\ &\leadsto r^{-1}s^{-2}r^4 \\ &\leadsto\dots\\ &\leadsto s^{-2}r^3.\end{align}$$ After we have reached this form, applying $r^n=1$ and $s^2=1$ repeatedly leads us to a uniquely determined word for each group element—this is called a normal form. Counting possibilities for the exponents of $r$ and $s$ leads to a total of $n\cdot 2 = 2n$ possible words (the normal form for the identity element is $s^0r^0$), thus yielding an upper bound on the size of any group satisfying these relations, i.e. $|G| \le 2n$.
The fact that there is a group of order $2n$ satisfying these relations, the dihedral group, tells us that the dihedral group has a presentation given by $G$.
In general, an argument just from a group presentation itself is much more likely to give you an upper bound on the order of the group than a lower bound, simply because it's rather tough to argue that, given a group presentation, a word cannot represent the identity. Indeed, there's a famous open conjecture about presentations of the trivial group!