Find a presentation with generators $g = (1,2)$ and $h = (2, 3)$ and three relations. 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.
 A: (a) To prove your presentation is actually a presentation of $S_3$, you have to do three things:


*

*Prove the generators in your presentation actually obey your relations. (This shows that the group presented by your presentation has a homomorphism to $S_3$.)

*Prove they actually generate the group. (This step shows that the homomorphism to $S_3$ is surjective.)

*Prove your set of relations is complete. (This step shows that the homomorphism to $S_3$ is actually an isomorphism. This is equivalent to proving that $\ker F_S\rightarrow G$ is equal to $\langle R\rangle^{F_S}$, but I'm personally unaware of a way to accomplish this step that works directly with the free group.)


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.
A: It's clear that $g$ and $h$ are elements of $S_3$ that satisfy the relations you gave.  
You still need to show that every element of $S_3$ is a word in $g$ and $h$.  That is, that $S_3=\langle g,h\rangle$.
But since $gh=(123)$, that gives you all three-cycles.
Now you are only missing one element: $(13)$.  But $(13)=(12)(132)$.
As @Ben Blum-Smith points out, it remains to show there are at most $6$ elements.
A: The easiest way to  see that this is actually a complete presentation of the  group comes from the following argument. Let $G$ be a group with presentation of the form
$$ G = \langle a,b \mid a^2 = b^2 = (ab)^n = 1 \rangle, \text{where }n\ge 1. $$
We will prove that $G$ is finite by showing that it is a quotient of the dihedral group  $D_n = C_n\rtimes C_2$. Indeed, the elements $ab$ and $a$ certainly generate $G$, satisfy $(ab)^n =  a^2 = 1$, and we have
$$a^{-1}(ab)a = ba = b^{-1}a^{-1} = (ab)^{-1}. $$
Therefore there is a surjective homomorphism  $C_n\rtimes C_2 \to G$, proving that $G$  has at most $2n$ elements. In fact it's not hard to show that $G$ is isomorphic  to the dihedral group, but I'll  leave  that to you.
A: Assume that the Free group $F:=F_S$ is generated by a set $S$.  Then every element of the group can be written in one and only one way as a product of finitely many elements of $S$ and their inverses. In particular, no relation gives you the identity element except identity. Now let $R$ be a subset of $S$. Let $N:= \langle R \rangle^F$ be the normal closure of $R$ in $F$ (it is the intersection of all normal subgroups of $F$, containing $R$). We put $G = F /N$. The element of the group $G$ are of type $\bar{x}= xN$, where $x \in F$. Now consider the natural projection map $F \to G$ with $x \mapsto \bar{x}$. The kernel of this map is $N$, so we have $G \cong F/N$. Therefore, the set of relations could be any set whose normal closure equals $N$. Notice that if we choose a relation $r \in R \subseteq N$, then $rN$ would be an identity element of $G$. This answers your question $b$. 
