Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 141 / 3 / c / Solution

Keep all four relators and all five generators, in the orders $(r_A,r_B,r_C,r_D)$ and $(d,e,a,b,c)$. Under abelianization, these generators become $(t,t,t^2,t^3,t^2)$. The Fox calculus rules areFor example, $r_B=a{d}^{-1}c{b}^{-1}$ gives the nonzero Fox derivative entries $-a{d}^{-1}$, $1$, $-a{d}^{-1}c{b}^{-1}$, $a{d}^{-1}$ in columns $d,a,b,c$, respectively. After abelianization these are $-t,1,-1,t$. The full Alexander matrix isThe Fox calculus identity supplies a useful check:Deleting the $e$ column, adjacent to the unbounded region across the marked meridian of a knot, givesThus $\det B$ is a representative of the Alexander polynomial of a knot of the figure-eight knot, with its customary ambiguity of multiplication by $\pm t^n$.