# Alexander Invariant of Torus knot

I am very interested in knot theory, especially in knot groups and knot polynomials. As is well known, it is easy to calculate the Alexander polynomial from the fundamental group $\pi_{1}(K)$ of a knot $K$ by free calculus. But I now am reading the book of Rolfsen (Knots and Links) which gets the polynomial by calculating the Alexander invariant from $\pi_{1}(K)$. I want to compute the Alexander polynomial of the torus knot $T_{p,q}$ for $p$ and $q$ coprime by the method in Rolfsen's book. There is a hint in his book as following:

• The knot group has presentation $G(T_{p,q})=( u,v\mid u^p=v^{q})$ where $u\mapsto q,v\mapsto p$ under abelianization.

• Choose integer $r,s$ satisfying $pr+qs=1,r>0,s<0$.Let $x=u^{s}v^{r},a=ux^{-q},b=vx^{-p}$ to obstain the presentation with $x\mapsto 1,a\mapsto 0,b\mapsto 0$:

$$G(T_{p,q})=(x,a,b\mid (ax^{q})^p=(bx^p)^{-q},x=(ax^q)^s(bx^p)^r)$$

• Let $C=[G,G]$ then $C/[C,C]$ has a $\Lambda-$module presentation with generators $\alpha,\beta$ and relations:

$$(t^q+t^{2q}+...+t^{pq})\alpha=(t^p+t^{2p}+...+t^{qp})\beta$$

$$(t^q+t^{2q}+...+t^{(-s)q})\alpha=(t^p+t^{2p}+...+t^{rp})\beta$$

• $H_1(\tilde{X})\cong \Lambda/(\Delta(t))$ where

$$\Delta(t)=\frac{(t^{pq}-1)(t-1)}{(t^p-1)(t^q-1)}$$

I know the (1)-(3),but I do not know how to get the (4) from (1)-(3). I need to know $\beta=(?)\alpha$ by eliminating the generator $\beta$ from the two relations.Can someone help me with this? Thanks a lot.

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Convert the matrix to its Smith normal form. In this particular case, this is doable over the integers. –  Alex Degtyarev May 10 '14 at 5:56
Just a remark: torus knot complements are mapping tori of finite-order automorphisms, from which one may deduce the Alexander polynomial by taking the characteristic polynomial. –  Ian Agol May 10 '14 at 17:36
–  Dietrich Burde May 10 '14 at 18:19
I want to get the relation $\Delta(t)\alpha=0$ by eliminate the gennerator $\beta$.So we get the $\Lambda-$module $H_1(\tilde{X})\cong(\alpha\mid \Delta(t)\alpha=0)\cong \Lambda/(\Delta(t))$ –  Jacob.Z.Lee May 11 '14 at 1:19
yeah. We can get the result by other methods ,like free calculus . –  Jacob.Z.Lee May 11 '14 at 1:26