How to compute the Alexander polynomial of a torus knot with Fox free calculus 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!
my calculations
 A: 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.
