Alexanderpolynomial of connected sum via Fox calculus and Wirtinger presentation Hello :) i have just reading the question "How to compute the Alexander polynomial of general torus knot" and i was suprised how strong it works if someone have a difficult question. I am also very interested in kont theory, very wonderful topic, but sometimes there are strong questions. 
There are many definitions how to compute the alexanderpolynomial (Fox calculus, via Dehn presentation, via Wirtinger presentation and other combinatorical definition - these are all equivalent definitions (but thats another topic to ask :D). My question is about the connected sum of to knots and the Alexanderpolynomial belongs to de connected sum.
I can prove with the definition via the Seifertmatrix that $$\Delta_{K_1\oplus K_2}(t)=\Delta_{K_1}\Delta_{K_2}$$ Thats not so difficult because the loops on the Seifertsurfaces are disjoint. But now i want to to it via the Wirtinger presentation and Fox calculus. Surely you may find the same result, but how to make a conclusion over the Wirtinger presentation of the connected sum?! Is it just the addition of the relations of how to handle this problem?? Can someone help me with this question? Thank you!
 A: Yes, this approach works.
Take a diagram demonstrating the connect sum (so that there is some circle intersecting $K_1+K_2$ in two points bounding $K_1$ on one side of the circle and $K_2$ on the other).  The Wirtinger presentation has an interesting property: a meridian generator for one of the two strands intersecting the circle is equal to the meridian generator for the other strand.  In the following picture, the green curves are the equal meridian generators; the basepoint is directly between the strands.

This means if $G_i=\pi_1(S^3-K_i)$ are the knot groups and if $a_i\in G_i$ are the meridian generators for the arcs of the connect sum, $\pi_1(G_i-(K_1+K_2))\cong \langle G_1,G_2\mid a_1=a_2\rangle$ (I'm using this as shorthand for "a group presentation given by the presentations of $G_1$ and $G_2$ along with the additional relation $a_1=a_2$").
Simplify the presentation by substituting $a_1$ for $a_2$ in $G_2$, then after taking the Fox derivatives and crossing off the row/column for $a_1$, the presentation matrix will be in block diagonal form, with each block corresponding to $G_1,G_2$.  The determinant of a block diagonal matrix is the product of determinants, so you have multiplicativity of the Alexander polynomial.
