Knot groups with big number of generators I start by saying that I am not an expert in this field and I apologize if the question is too elementary.
Let $K$ be a knot in $S^3$. I denote by $\pi_1(K)$ the knot group, which is the fundamental group of its exterior:
$$ \pi_1(K) = \pi_1(S^3 \smallsetminus K) .$$
The minimal number of generators of a knot $K$ is the minimal number of generators of $\pi_1(K)$.
I am searching for knots with an arbitrarily high minimal number of generators. In particular:

*

*I found in https://arxiv.org/abs/1007.3175 , Lemma 4.17, a reference: Goodrick, R. (1968). Non-simplicially collapsible triangulations of In. In this article, the author proves that the connected sum of $n$ copies of a two-bridge knot is a $m$-bridge knot with $m$>$n$. A sharper result should hold from Schultens, Jennifer (2003). Additivity of bridge numbers of knots. I cannot understand, though, how this should prove the statement. In particular, by Knot theory question: bridge number vs. min generators of fundamental group of complement , this should not imply the result. Maybe I am missing something in the article, I did not go through the details. However, the article is quite dated and there is, hopefully, a simple way to prove this fact nowadays. So the first question is: is there a simple way to prove that there are knots with an arbitrarily high minimal number of generators?


*The techniques used seem to rely on the connected sum. What if we search for prime knots with an arbitrarily high minimal number of generators?


*What if we search for hyperbolic knots with an arbitrarily high minimal number of generators?
Thank you in advance for the attention.
 A: I think the rank of the Alexander polynomial gives you a lower bound on the number of generators of your fundamental group.  i.e. just compute the Alexander module by lifting a 2-complex for the knot exterior to the universal cover.
It's a very coarse bound, but it seems to answer your question, as there are hyperbolic knots with arbitrarily large Alexander polynomials.
edit: my estimation was too quick.  See Kyle Miller's comment below.  His argument gets a lower bound and carefully avoids my mistake. But the statement needs to be changed, i.e. he uses full Alexander modules rather than Alexander polynomials.
A: If $\pi_1(S^3\setminus K)$ has a presentation with $n$ generators then its representation variety $\mathrm{Hom}(\pi_1(S^3\setminus K),SL_2(\mathbb{C}))$ is a subvariety of $(SL_2(\mathbb{C}))^n$, which has complex dimension $3n$, so any component will have complex dimension at most $3n$.    So if you want the minimal number of generators to be arbitrarily large, you just have to find knots with high-dimensional representation varieties.  (You can replace $SL_2(\mathbb{C})$ with other groups if you prefer.)
Cooper and Long ("Remarks on the A-polynomial of a knot", section 8) show how to construct hyperbolic knots where there are components of arbitrarily large dimension.  Start with some $n$-fold connected sum, and realize it as the closure of a braid $\beta$ such that the union of the closure $\hat\beta$ with its braid axis is hyperbolic.  Lift $\hat\beta$ to the $p$-fold cyclic branched cover of $S^3$, branched over the braid axis -- this will again be $S^3$, since the axis is unknotted -- to get a new knot, which is hyperbolic for all large enough $p$.  (This knot will be the closure of the braid $\beta^p$.)  Then the representation variety of this new knot turns out to have a component of dimension at least $n$ as well.
