I would like to calculate the Alexander polynomial of a torus knot with presentation $\langle x,y \mid x^p=y^q \rangle$ using Fox free calculus. I would also like to get to $\frac{(t^{pq}-1)(t-1)}{(t^p-1)(t^q-1)}$.
Below are my calculations; the Alexander matrix calculation is at the end. I can't see my mistake. I thought that with just two generators it doesn't matter which of the two partial derivatives I calculate because the Alexander polynomial is just one of the two $1 \times 1$ matrices of the Alexander matrix. I also looked in Rolfsen's Knots and Links and Burde and Zieschang's Knots. The latter states that the Alexander matrix is:
$$\begin{pmatrix} \frac{t^{pq}-1}{t^q-1} & - \frac{t^{pq}-1}{t^p-1} \end{pmatrix}$$
It also states that the greatest common divisor here is the Alexander polynomial. I thought it was just one of the two entries for two generators.
But I don't even get to this Alexander matrix. In my calculations the $\alpha$ is the Abelianization and $\gamma$ is the canonic homomorphism. I'm a little confused and hope somebody can help!
The presentation matrix that comes from using the Wirtinger presentation is fairly special, and in general you do have to calculate the GCD of all the appropriately sized minors.
The $(p,q)$ torus knot has the efficient presentation $\langle x,y\mid x^p=y^q\rangle$. Let me reproduce the Fox derivatives and such. First, the abelianization $\alpha$ can be defined by $x\mapsto t^q$ and $y\mapsto t^p$, which one can get from knowing that $\alpha(x)^p=\alpha(y)^q$ and that $\gcd(p,q)=1$ (it follows from these facts, too, that the abelianization has one generator, which I am calling $t$). \begin{align*} \frac{\partial x^py^{-q}}{\partial x}&=1+x+x^2+\dots+x^{p-1}\\ \frac{\partial x^py^{-q}}{\partial y}&=x^p\frac{\partial y^{-q}}{\partial y}=x^p(-y^{-1}-y^{-2}-\cdots-y^{-q}) \end{align*} There is no need to use any properties of the canonical homomorphism directly, and we can pretend that $\alpha$ is pulled back to the free group: applying the abelianization, we have \begin{align*} \alpha\frac{\partial x^py^{-q}}{\partial x}&=1+t^q+t^{2q}+\dots+t^{(p-1)q}=\frac{1-t^{pq}}{1-t^q}\\ \alpha\frac{\partial x^py^{-q}}{\partial y}&=t^{pq}(-t^{-p}-t^{-2p}-\dots-t^{-pq})\\ &=-t^{p(q-1)}-t^{p(q-2)}-\dots-1=-\frac{1-t^{pq}}{1-t^p} \end{align*} Hence, we have the presentation matrix \begin{pmatrix} \frac{1-t^{pq}}{1-t^q} & -\frac{1-t^{pq}}{1-t^p} \end{pmatrix} The GCD of the $1\times 1$ minors is $\frac{(1-t^{pq})(1-t)}{(1-t^p)(1-t^q)}$, which you can get with a little knowledge of the properties of cyclotomic polynomials. This is the Alexander polynomial.
One mistake I noticed in your work was that you used $x\mapsto t$ and $y\mapsto t$ for the abelianization, which is not correct.