TITLE: Presentation of $D_{2n}$
QUESTION [10 upvotes]: The presentation of the dihedral group $D_{2n}$ is 
$$D_{2n}= \langle r,s \mid r^{n}=s^{2}=1, rs=sr^{-1} \rangle.$$
Why is it incorrect to conclude that $r=s=1$ ?   In other words why doesn't this group presentation describe the trivial group.

REPLY [2 votes]: In the question
"$r^{n}=s^{2}=1, rs=sr^{-1}$ ... why is it incorrect to conclude that $r=s=1$ ?  ... why doesn't this group presentation describe the trivial group"  
the key words are "conclude" and "describe".  You are in effect asking what logic applies to statements about $r$ and $s$.
The answer is that assertions about $r$ and $s$ are understood as valid in this context only if they hold in ALL groups.  The assertion "if a pair of group elements $r$ and $s$ satisfies the defining relations of $D_{2n}$, then $r=s=1$" is false in some groups and true in others.  It is true in groups with no subgroup of order $2n$, such as large cyclic groups of prime order.   It is false in other groups, such as the dihedral group of order 2n, or the group of order 2 where $r=1, s=-1$ is a pair satisfying the equations.
Truth in all groups is a semantic definition and can be replaced by an equivalent, syntactic definition that gives the combinatorial rules for making all valid deductions from the defining relations.  $r=s=1$ is not a valid algebraic consequence of the algebraic rules.   As in all problems of logic, showing that something does NOT follow from some rules or axioms is a matter of constructing an example (a model) where the rules apply but the assertion is false.  Here the models are groups with a pair of elements $r,s$ satisfying the relations, so for showing what is not true in the logic of generators and relations the algebraic approach doesn't provide an alternative to constructing models.  For assertions that do follow from the defining relations, the algebraic definition is useful, because instead of surveying a possibly infinite collection of models to show that the statement is correct, one can perform a finite combinatorial derivation that works for all models.
(Added in response to comment from Matt E about normal forms: showing that any element can be placed in the normal form follows from the algebraic relations alone.  However, to use normal forms as a method of calculating in the group or understanding it, you need to show that distinct elements in the group have different normal forms. This is again the non-algebraic (or not purely algebraic) problem of showing that relations do not hold in a structure, which is done by constructing models such as actions of the group on particular spaces.)