I am reviewing an exercise I did some time ago about the presentation of the symmetric group $S_3$ but I'm having doubts.
I was asked to find a presentation with generators $g = (1,2)$ and $h = (2, 3)$ and three relations.
The answer I gave was: $$\langle g, h\mid g^2 = h^2 = (gh)^3 = id \rangle.$$
Now, my questions are:
a) The person who corrected the exercise added that I should have proved that what I found is actually a presentation. How do I do this? I know that I should check that the kernel of the associated surjective homomorphism $F_S \to G$ is equal to $\langle R\rangle^{F_S}$ where $R$ is the set of our relations, but I don't get what this last part means. How do I proceed?
b) In general, if I'm asked to find a presentation, is it correct to look for the cases where the generators are equal to the identity element? (For example, from above we have that g squared is the identity). Otherwise, which criterion do I use to find relations?
I'm sorry if this is confusing but the fact is that I don't remember very well what all this means since some time has passed from the last time I saw these things.
Best Answer
(a) To prove your presentation is actually a presentation of $S_3$, you have to do three things:
Now, I assume you know how to show $g,h$ actually obey your stated relations. And Chris Custer discussed how to show they actually generate the group. These two steps tell us that there is a surjective homomorphism from the group presented by $\langle g,h\mid g^2=h^2=(gh)^3=id.\rangle$ onto $S_3$. The hard part is proving your relations are complete, i.e. that they generate all the relations between $g$ and $h$, which is equivalent to saying that this surjective homomorphism is also injective.
I assume that whoever graded your assignment and asked you to do this had some particular way in mind for you to go about it, and I can't guess what that is; but, I myself would do this with the Todd-Coxeter algorithm. (See here for more of an exposition.) This algorithm allows you to prove that the group presented by $\langle g,h\mid g^2=h^2=(gh)^3=id.\rangle$ has only 6 elements, and this would allow you to conclude that its surjective homomorphism onto $S_3$ is also injective.
(b) In general, if the group $G$ is finite, then there will be powers of the generators that equal the identity, and you can use these as relations; however, they won't in general be a complete set of relations. Even in your example, $g^2=h^2=id.$ isn't enough; you also needed to find $(gh)^3=1$, and $gh$ is not one of your generators. Finding a complete set of relations can be hard, depending on the form in which the group comes to you.
If the group $G$ isn't finite, then you don't even have a guarantee that your generators will have powers equal to the identity. You may have to look elsewhere for relations.
On the other hand, certain types of groups promise you certain forms for the presentation. For example, if the group is finite and solvable (as $S_3$ is), then a "power-conjugate presentation", also called a "polycyclic presentation" ("PC presentation" either way) is guaranteed to exist. See here which has a clear explanation. On the other hand, the $g$ and $h$ of the OP are not a polycyclic sequence for $S_3$, so this wouldn't have helped you here.
Finding a presentation of an arbitrary group that comes to you in an arbitrary form isn't a straightforward task, but I assume that whenever you are asked to find a presentation in a coursework context, there will be specific features of the group (e.g., for $S_3$, its small size) that give you some leverage.