The fundamental group of a knot $K$ (otherwise known as the knot group) is the fundamental group of the knot complement $S^{3} \backslash K $ in $S^{3} $.


In "Virtual Knots: The State of the Art" (<http://books.google.com.au/books?id=WaCJ_-MpdBYC>) on page 7, it says that the fundamental group of a knot recognizes prime knots. Does this mean that the knot group of a prime knot cannot be isomorphic to the amalgamated free product of two non-trivial knot groups- and in general can the knot group of a prime knot be an amalgamated free product?



The knot group of a prime knot
The statement on page 7 is referring to Corollary 2.1 of Gordon and
Luecke's "[Knots are determined by their complements](http://www.ams.org/journals/jams/1989-02-02/S0894-0347-1989-0965210-7/S0894-0347-1989-0965210-7.pdf)."


Also a prime knot can not be an amalgamated free product of
non-trivial knot groups by Corollay 2.5 of W. Thurston's "[Three
dimensional manifolds, Kleinian groups and hyperbolic geometry](http://www.ams.org/journals/bull/1982-06-03/S0273-0979-1982-15003-0/S0273-0979-1982-15003-0.pdf)."


However, a knot group can be an amalgamated free product provided the
subgroup you are amalgamating along is more complicated. For example,
a knot complement with an essential, embedded quasi-Fuchsian surface
has such a knot group. The basic idea of how to construct such knots
goes as follows, take a genus $g$ handlebody and embed a knot in it
such that the knot punctures any compressing disk in at least two
points, then embedded the handlebody in $S^3$ such that the boundary
of the handlebody is incompressible. This process is done in full
rigor by Adams and Reid in "[Quasi-Fuchsian surfaces in hyperbolic knot complements](http://dx.doi.org/10.1017/S1446788700031967)."


