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The knot group of a knot $K$ is the fundamental group of the knot complement $\pi_{1} (S^{3} \backslash K )$ (sometimes denoted $\pi_{1} (K)$ ).

Let $f: \pi_{1} (K) \rightarrow \pi_{1} (L) $ be a knot group epimorphism and suppose that $K$ is a prime knot. Can $L$ be a composite knot?

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2 Answers 2

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The answer to this question is yes, L can be a composite knot.

Here is an outline of how to achieve such an epimorphism:

If a knot complement $S^3-K$ is periodic, i.e. there is symmetry $\tau$ of the knot in that fixes points in $S^3$, then the knot complement covers a knot complement in an orbifold $Q$ where the base space of $Q$ is $S^3$. In this case, $Q-K' \cong S^3-K/\langle \tau \rangle$ and the base space of $Q-K'$ is some knot complement. To agree with the notation of your question, let's say this base space is $S^3-L$.

Meridians of $\pi_1(S^3-K)$ map to $\pi_1(S^3-L)$ and the covering then base space operation is a degree 1 map, so there is an epimorphism $f:\pi_1(S^3-K) \rightarrow \pi_1(S^3-L)$.

In order, to construct $Q-K'$, we can take a two component link $K'\cup C$ where $K'$ is a composite knot and $C$ is some unknotted component with non-zero linking number $\ell$ with $K'$ and $S^3-(K' \cup C)$ is hyperbolic.

Here is an example of such a $K'\cup C$.

Since $S^3-(K' \cup C)$ is hyperbolic, there are infinitely many $(n,0)$ fillings on $C$ that are hyperbolic with $n$ and $\ell$ relatively prime. The last condition assures that $S^3-(K' \cup C) (n,0)$ is covered by a knot complement and choosing $n$ to be sufficiently large assures there is an $S^3-K$ that covers $S^3-(K' \cup C) (n,0)$ is hyperbolic and hence $K$ is prime. However, $S^3-K'$ could be composite as in the example.

(I used snappy to give evidence of hyperbolicity for this example. Strictly speaking this is not a proof of primeness. However, (4,0) surgery seems to make this example hyperbolic, i.e. the filling has 'solution type all positively oriented tetrahedra'.)

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This is a very nice answer (and picture !). –  Dietrich Burde Jul 10 '13 at 14:35

Yes, in general $L$ need not be a prime knot, if $K$ is a prime knot, given a knot epimorphism $f\colon \pi K \to \pi L$, see section $4$ of D.S. Silver and W. Whitten, Knot group epimorphisms II (which is referring to part I for an example).
There it is also discussed which properties of $K$ are inherited by $L$, given a knot group epimorphisms $f\colon \pi K \to \pi L$. For example, if $K$ is alternating, $L$ need not be alternating in general. On the other hand, if, say, $K$ is a torus knot, then $L$ is, too.

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Thanks Prof. Burde :) –  Peter Jul 11 '13 at 2:48

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