# Is a group defined by its generator set and relations?

I'm learning about generators from Dummit and Foote. They call this a presentation of the dihedral group:

$$D_{2n} = \left< r,s\,|\, r^n=s^2=1,\, rs=sr^{-1}\right>$$

Does this type of "presentation" determine/define a group? I think there are several groups that fulfill the above relations, e.g. the trivial group with $r=s=1$. An exercise is to determine the order of a group given such a "presentation". Is there some assumption I'm missing? Could someone clear this up for me?

• The key point you are missing is that the stated relations are the only relations that the generators satisfy. – Michael Albanese Jan 11 '16 at 9:46
• You assume this to be the minimal set of relations. – Paul K Jan 11 '16 at 9:47
• For that exercise, you can use Todd-Coxeter Al. to find the orders of some certain finite groups. It really works! – mrs Jan 11 '16 at 9:57

So, the trivial group does satisfy the above relations, but it also satisfies the relation $s=1$, a relation that does not follow from the above relations, meaning it is not the group represented in this particular case.
More technically, what you wrote down as a definition of $D_{2n}$ is in fact a shortened notation, meaning that $D_{2n}$ is isomorphic to $G/H$ where $G$ is a free group with 2 generators and $H$ is the normal subgroup generated by all elements of the type $s^2$, $r^ns^{-2}$ and $rsrs^{-1}$ (i.e., all elements that are equal to $1$ in $D_{2n}$)
Using this, you can show that any group that satisfies the listed properties is a quotient of the generated group. If it satisfies only the listed properties, it is the whole group, if it satisfies some other properties (like $s=1$, it is a proper quotient (and, of course, the trivial group is a proper quotient of $D_{2n}$)