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?

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

The idea of presentation is that the elements of the group are allowed to have the listed properties, properties that follow from them, and no other properties.

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 $D_{2n}$)

  • $\begingroup$ @Anna I added a more strict definition as well. $\endgroup$ – 5xum Jan 11 '16 at 9:51
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    $\begingroup$ In the last paragraph you want “quotient”, not subgroup. $\endgroup$ – Carsten S Jan 11 '16 at 11:05

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