Alexander Polynomial of the stevedore knot using Fox's free calculus

Question

I want to find the Alexander Polynomial of the stevedore knot using the free calculus of Fox. A presentation for the knot group of the stevedore knot is

$$G=(x,a:xa^3xa^{-2}x^{-1}=a^2xa^{-2})$$

We can rewrite the relator as $r=a^2x^{-1}a^{-2}xa^3xa^{-2}x^{-1}=1$. To find the Alexander matrix we find the free derivatives of the relator with respect to each generator:

$$\frac{\partial r}{\partial a}=1+a-a^2x^{-1}(a^{-1}+a^{-2})+a^2x^{-1}a^{-2}x(1+a+a^2)-a^2x^{-1}a^{-2}xa^3x(a^{-1}+a^{-2})$$ $$\frac{\partial r}{\partial x}=-a^2x^{-1}+a^2x^{-1}a^{-2}+a^2x^{-1}a^{-2}xa^3-a^2x^{-1}a^{-2}xa^3xa^{-2}x^{-1}$$

To abelianize these, map the generators $(a,x)\rightarrow t$. We get two polynomials,

$$1+t-t(t^{-1}+t^{-2})+(1+t+t^2)-t^4(t^{-1}+t^{-2})=1+2t-t^3-t^{-1}$$ $$-t+t^{-1}+t^3-t=-2t+t^{-1}+t^3$$

So the Alexander Matrix is

$$[t^{-1}(-t^4+2t^2+t-1)\quad t^{-1}(t^4-2t^2+1)]$$

The Alexander polynomial should be the generator of the ideal generated by all the minors of this matrix. Ignoring the prefactors, the ideal is

$$I=<-t^4+2t^2+t-1,t^4-2t^2+1>$$

But, the Alexander polynomial of the stevedore knot is $2t^2-5t+2$. Since this doesn't divide my two generators, it looks like I've done something wrong. Can someone set me straight?

(Note: This question is an attempt to rephrase this question to get more attention and maybe isolate the key features of. I would consider an answer here to be an answer there and will remove it if this works)

Answer 1

Ok, I've worked this out so I'll post the answer here and close the associated question.

The issue is specific to the stevedore knot. In the presentation given above, the generators are listed as $a$ and $x$, but the generators of the under/overcrossing edges are $x$ and $ax$. So the representation is not the Wirtinger one - it's Wirtinger + the identity $y=ax$, where $y$ is the generator actually passing under one of the edges. If you rewrite the relator using $y$ instead of $ax$, you reproduce the correct Alexander polynomial.

I'm sure this is somehow related to the fact that the Alexander polynomial is at some level related to the crossings in the diagram, whereas this particular presentation of the group has been simplified by adding an extra relation.