TITLE: Semidirect product decomposition of the Borromean rings group
QUESTION [15 upvotes]: Let $X=S^3\setminus B$ be the link complement of the Borromean rings.
    (source)
Then $G=\pi_1(X)$ has a presentation of the form
$$
G = \langle \; a,b,c \mid [a,[b^{-1},c]],\; [b,[c^{-1},a]], \; [c,[a^{-1},b]]\;\rangle,
$$
where any one of the relations is redundant.
Removing any component of the link results in a trivial link of two components. The effect on the fundamental group is to send one of the generators, say $c$, to $1$, resulting in a homomorphism $G\to F_2$ to the free group on $2$ generators. There results a split extension
$$
1\to K\to G\to F_2\to 1.
$$

Is the kernel $K$ a free group?

REPLY [5 votes]: I believe that $K$ is an infinitely generated free group.  First we find an infinite presentation $K = \langle X \mid R \rangle$, and then we argue that we can pick an infinite subset of $X$ that freely generates.
Presentation: Note that we can obtain the Borromean rings from the handlebody of genus two by drilling out the commutator curve and attaching a two-handle to the boundary along the usual separating curve.  Since the attaching curve for the two-handle is trivial in the handlebody, we can take the universal cover of the handlebody and then attach infinitely many two-handles, and drill infinitely many lines.  After drawing a few pictures, we can organize all of this information as follows. 
Let $D^2$ be the Poincare disk model for the hyperbolic plane.  Pick hyperbolic isometries $a$ and $b$ that have perpendicular axes and equal (very long) translation lengths.  Thus $F_2 = \langle a,b \rangle$ is a free group of rank two.  Let $T$ be the regular four-valent tree isometrically embedded in $H^2$, arising as the Cayley graph of $F_2$.  Let $v_0$ be the vertex of $T$ corresponding to the identity of $F_2$.  The universal cover of the handlebody above is obtained as a three-dimensional neighborhood of $T$, and $F_2$ is the deck group. 
We define $X = \{ x_q \}$ where $q$ ranges over the connected components of $H^2 - T$.  (These correspond to the drilled lines.)  This gives the generating set for $K$ subject to the relations $R = \{ x_p X_q x_r X_s \}$ where the regions $p,q,r,s$ are arranged counter-clockwise about a vertex of $T$. (These correspond to the attached two-handles.  Because we started with the commutator curve, the two-handles are attached to the neighborhood of $T$ via a "baseball curve" around each vertex.)  Here we use the convention that $X_q = (x_q)^{-1}$.  (Also, $p$ is always to the "north-east" of $v$ - we use translates of the picture about $v_0$ to set conventions.)  For future use, let $w_v$ be the relation coming from vertex $v$. 
Subset that freely generates:  We recursively color the regions of $H^2 - T$ as follows.  (1) All regions start white. (2) In the first step, we pick two regions $p, q$ adjacent to $v_0$.  Color both of $p$ and $q$ black.  (3) In general, pick a vertex $v$ having exactly two white regions $p, q$ adjacent to $v$.  Note that $p$ and $q$ must be adjacent to each other.  Color $p$ black and color $q$ grey. 
The set of black regions gives a subset $X' \subset X$ of the generators.  It is an exercise to prove inductively that $X'$ generates.  Next, for a contradiction, suppose that $w$ is a word in the $X'$ that is trivial in $K$.  Thus we can write $w$ as a product of conjugates of relations from $R$.  Recall that relations are in bijection with the vertices of $T$.  Pick one relation $w_v$ appearing in the product so that $v$ is as far as possible from $v_0$.  Of the four regions adjacent to $v$, there are two regions $p, q$ that are combinatorially further than the others from $v_0$.  Say $p$ and black and $q$ is grey.  Then $x_q$ is not freely cancelled in the product of conjugates, a contradiction. QED