For $X = S^3 - K$ a knot complement, the abelianization $$1 \to [\pi_1(X),\pi_1(X)] \to \pi_1(X) \to \mathbb{Z} \to 1$$ has a right split, where $\mathbb{Z} \cong \pi_1(X)^{ab}$ is realized by a meridian loop. This yields the semidirect product $\pi_1(X) \cong [\pi_1,\pi_1] \rtimes \mathbb{Z}$.
My question is: how do I see the commutator subgroup geometrically? I.e. the $\mathbb{Z}$ subgroup is a meridian loop as I said. Is there some similar geometric description of the commutator subgroup?
When $X$ is fibered, the infinite cyclic cover, which is the cover corresponding to $[\pi_1,\pi_1] \leq \pi_1(X)$, deformation retracts to the fiber surface (I think). So the commutator subgroup is just the fundamental group of the fiber surface (which in particular is a free group). Can we say anything when the knot is not fibered? Like maybe you get the fundamental group of a minimal genus Seifert surface?
Edit: according to a theorem of Stallings (1962), the subgroup $[\pi_1,\pi_1]$ is finitely generated if and only if $K$ is fibered. So in the non-fibered case, this subgroup is going to be pretty wild. Is there anything we can still understand about it? Is it still a free group?