TITLE: Finding the kernel of an epimorphism onto $S_3$?
QUESTION [6 upvotes]: Let $\Lambda$ denote the group with presentation $\langle a,b \mid 
 abab^{-1}a^{-1}b^{-1}\rangle$. We define the following epimorphism
  from $\Lambda$ onto $S_3$:
$\theta: \Lambda(a,b) \rightarrow S_3$ using $a \mapsto (12)$ and $b
 \mapsto (23)$. We have $(12)(23)(23)^{-1}(12)^{-1}(23)^{-1}=e$
  thus indicating that our homomorphism factors through the quotient
  modulo the normal subgroup $H$ of $F$ generated by
  $abab^{-1}a^{-1}b^{-1}$ and, by definition, $\Lambda=F/H$. Find an
  explicit, finite presentation for the kernel, $\kappa$, of the
  epimorphism

I know the kernel of $\theta$ is the set of all elements of $\Lambda$ that are mapped to $e\in S_3$ and is a normal subgroup of $\Lambda$. I'm a little new to finding kernels so I'm not sure how to proceed. What's a good way to find the kernel of this epimorphism?

REPLY [2 votes]: There is an algorithm known as the Reidemeister-Schreier algorithm for computing presentations of subgroups of finitely presented groups. I haven't got time to explain it right now and in any case you might do better to google it - there are lots of descriptions online.
Examples like this, where the index of the subgroup is small, can easily be done by hand. For subgroups of larger index, you can sue GAP or Magma. I was lazy and did your example by computer and found the presentation of the kernel:
$$\langle x,y,z \mid y^{-1}x^{-1}yzxz^{-1}, y^{-1}zxyx^{-1}z^{-1}  \rangle,$$
where $x=a^2$, $y=b^2$, and $z=ab^2a^{-1}$.