= Solution
The <Wirtinger presentation> assigns a meridian generator to each arc between undercrossings. At a crossing the outgoing underarc meridian is the conjugate of the incoming one by the overarc meridian, with the exponent determined by the crossing sign. One crossing relation is redundant for a connected <knot> diagram. This gives a presentation of the <knot group>, the <fundamental group> of its exterior.
For the <trefoil knot>, choose three arcs x,y,z so that two crossing relations are $z=xyx^{-1}$ and $x=yzy^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, hence
$$
\boxed{\pi_1(S^3\setminus T)=\langle x,y\mid xyx=yxy\rangle.}
$$
We use the relator $r=xyxy^{-1}x^{-1}y^{-1}$.
Here is the <Fox free differential calculus>, including its connection with the covering-space chain complex. Let F be the <free group> on $x_1,\ldots,x_n$, let $\mathbb ZF$ be its <group ring>, and let $\varepsilon:\mathbb ZF\to\mathbb Z$ be augmentation. Define additive maps $D_j:\mathbb ZF\to\mathbb ZF$ by
$$
D_j(x_i)=\delta_{ij},\qquad D_j(ab)=D_j(a)\varepsilon(b)+aD_j(b),\qquad D_j(1)=0.
$$
For <group> words u,v the product rule reads $D_j(uv)=D_j(u)+uD_j(v)$. Differentiating $x_ix_i^{-1}=1$ forces $D_j(x_i^{-1})=-\delta_{ij}x_i^{-1}$. Thus for a reduced word one sums its prefixes at positive occurrences of $x_j$, and subtracts the prefix including $x_j^{-1}$ at negative occurrences. Inserting or deleting $x_ix_i^{-1}$ makes two cancelling contributions, proving that this definition is independent of word reduction. Linear extension proves the group-ring product rule and uniqueness.
The fundamental identity is
$$
w-1=\sum_{j=1}^nD_j(w)(x_j-1).
$$
It holds on generators and inverses. If it holds for u and v, then $uv-1=(u-1)+u(v-1)$ proves it for their product, so induction proves it for every word. These <Fox derivatives> form the <matrix> $\big(D_jr_i\big)_{i,j}$ for any chosen presentation.
Let $\phi:G\to\mathbb Z$ be knot-group <abelianization> and put $a_j=\phi(x_j)$. Apply the ring map $\alpha:\mathbb ZF\to\Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha(D_jr_i)$. Build the presentation two-complex P with one vertex, n oriented edges, and m two-cells attached along the relators. Its <infinite cyclic cover> has <cellular chain complex>
$$
\Lambda^m\xrightarrow{\,J^T\,}\Lambda^n\xrightarrow{\,d_1\,}\Lambda,\qquad d_1(e_j)=t^{a_j}-1.
$$
To verify the second boundary, lift an attaching word starting at deck level zero. An occurrence $x_j$ after prefix u traverses the jth edge at level $\phi(u)$, contributing $t^{\phi(u)}e_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\phi(ux_j^{-1})$, contributing $-t^{\phi(ux_j^{-1})}e_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha(r_i)=1$, gives $d_1J^T=0$.
The cover of P and the cover of the <knot exterior> both have <fundamental group> $\ker\phi$. Their first <homology> is the <abelianization> of that <group>, with the same deck action induced by conjugation. Thus no assumption that the presentation complex is aspherical is needed, and the complete relation with the <Alexander module of a knot> is
$$
\boxed{\mathcal A_K\cong\ker d_1/\operatorname{im}J^T.}
$$
For meridian generators all $a_j=1$. Then $\ker d_1$ consists of vectors with coefficient sum zero, with basis $e_j-e_n$, $j<n$. Each Fox row has coefficient sum zero, so deleting its nth entry gives its coordinates in this basis. Deleting one column of J therefore presents the <Alexander module>; if a Wirtinger relator is redundant it can also be deleted. This proves the <matrix> procedure used for the displayed <knot> above. For arbitrary generators, the exponent vector is primitive because the generators map onto $\mathbb Z$; elementary changes of free generators implement the Euclidean algorithm and reduce it to $(1,0,\ldots,0)$. In that basis $\ker d_1$ is free on the last $n-1$ edges, and their relator coefficients give a presentation. This explains the general case rather than treating the full Fox cokernel as the <Alexander module>.
For the trefoil relator, with $\alpha(x)=\alpha(y)=t$, direct differentiation gives
$$
\alpha(D_xr)=1-t+t^2=f(t),\qquad\alpha(D_yr)=t-t^2-1=-f(t).
$$
The first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whence
$$
\boxed{\mathcal A_T\cong\Lambda/(t^2-t+1),\qquad\Delta_T(t)\doteq t^2-t+1.}
$$
For a <connected sum of knots>, join <Seifert surfaces> in separated balls. Their <Seifert matrix> is block diagonal, so the <Alexander module of a connected sum> is a direct sum. Thus $\mathcal A_{T\#T}\cong(\Lambda/(f))^2$.
This <module> is not cyclic. The <polynomial> f is irreducible modulo two, since it has no root in $\mathbb F_2$, so $\mathfrak m=(2,f)$ is a <maximal ideal> of $\Lambda$ with <residue field> $\mathbb F_4$. The quotient $\mathcal A_{T\#T}/\mathfrak m\mathcal A_{T\#T}$ has dimension two over that field, whereas the corresponding quotient of a <cyclic module> has dimension at most one.
Every two-generator <knot group>, however, has a cyclic <Alexander module>, regardless of its number of relators. Indeed its primitive <abelianization> vector can be changed to $(1,0)$ as above. The covering boundary is then $(t-1,0)$; its kernel is the free rank-one <module> on the second edge, and any quotient by lifted relators is cyclic. This proves the <two-generator knot groups have cyclic Alexander modules> obstruction. Consequently \b[the <group> of the connected sum of two trefoils cannot even be generated by two elements], and in particular it has no two-generator, one-relator presentation.
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