Write $o(\theta )$ for the order of the determinant character of an irreducible character $\theta$. An extension has the same degree and restricts exactly to $\theta$, rather than just containing it as a constituent.

Extension theorem. Thompson's coprime character extension theorem says that if $N◃G$, $\theta \in \mathrm{Irr}(N)$ is $G$-invariant, and $\gcd (\theta (1),[G:N])=1$, then $\theta$ extends to $G$ exactly when its determinant character extends to a linear character of $G$. In particular, if also $\gcd (o(\theta ),[G:N])=1$, there is a unique extension with determinant order coprime to the index. For a p-group quotient, the version used below gives a canonical extension whenever both the degree and the determinant order are prime to $p$. This theorem is an assertion of extendibility, not the false statement that every invariant irreducible character extends.

The p-nilpotency lemma. Definewhere $m_p$ is the largest power of $p$ dividing an integer $m$. Isaacs' character-degree criterion for p-nilpotency assertsEquivalently, $s_p(G{)}_p=h_p(G)$. We prove both the divisibility and the criterion.

Let $N=O^p(G)$ be the p-residual of a finite group, the least normal subgroup with p-group quotient. It can also be described as the subgroup generated by all p-prime-order elements: every homomorphism to a p-group kills those elements, and after quotienting by their generated subgroup the image of every element has p-power order, so the quotient is a p-group by Cauchy's theorem.

The residual is p-perfect, meaning $O^p(N)=N$. Indeed $O^p(N)$ is characteristic in $N$, hence normal in $G$, and $G/O^p(N)$ is an extension of the two p-groups $N/O^p(N)$ and $G/N$. Minimality of $N$ forces equality. Consequently $N/N^{\prime }$ has order prime to $p$, since otherwise it would have a quotient of order $p$. Every linear character of $N$, in particular every determinant character of $N$, therefore has order prime to $p$.

Let $\mathcal{T}$ consist of the $G$-invariant irreducible characters $\theta$ of $N$ of degree prime to $p$. Each extends to an irreducible character $\hat{\theta }$ of $G$ by the extension theorem. The irreducible characters of $G$ lying over $\theta$ are exactlyHere is the representation argument for this description. If a representation lies over the invariant $\theta$, its $N$-module is copies of the module for $\theta$. Identify it with that module tensored with its multiplicity space. Dividing the $G$-action by the fixed extending action on the first factor leaves an action on the multiplicity space, trivial on $N$. Irreducibility is equivalent to irreducibility of that quotient action, proving the displayed parametrization.

Every irreducible degree of the p-group $G/N$ is a power of $p$, since it divides the group order. Thus the p-prime-degree characters lying over $\theta$ are exactly those with $\nu$ linear, and there areof them. For the last equality, $G/N{G}^{\prime }$ is the largest abelian p-group quotient of $G$, hence the p-primary part of its abelianization. In particular all these characters have degree $\theta (1)$.

Conversely, for any $\chi \in \mathrm{Irr}(G)$ of p-prime degree, Clifford theory writes its restriction to $N$ as $e$ times the sum of a single orbit of irreducible characters. The orbit size divides $[G:N]$ and also divides $\chi (1)$, so it is one; the constituent $\theta$ is invariant and has p-prime degree. Hence $\theta \in \mathcal{T}$. The preceding extension parametrization now forces the quotient character to be linear. We have therefore obtained the exact identityThis proves the required divisibility without assuming p-nilpotency.

The p-group $G/N$ acts on $\mathrm{Irr}(N)$ by conjugation. Every nonfixed orbit has size divisible by $p$, and degrees are constant on each orbit. Fixed characters whose degrees are divisible by $p$ also contribute zero modulo $p$. The sum of squares of irreducible degrees consequently givesThe quotient is prime to $p$ exactly when $N$ has order prime to $p$. Since $G/N$ is a p-group, that is exactly the assertion that $N$ is a normal p-complement. Conversely, any normal p-complement equals $O^p(G)$ because it contains all p-prime-order elements and has p-group quotient. This finishes the lemma.

Detection from the group algebra. For finite groups $G$ and $H$ with isomorphic complex group algebras, the Artin–Wedderburn theorem givesThe simple matrix-block sizes and their multiplicities are determined by the algebra isomorphism. Hence $G$ and $H$ have identical irreducible degree multisets. They have the same order, from the degree-square sum, and the same number of linear characters, which equals $|G:G^{\prime }|$. Thus $s_p$ and $h_p$ coincide for every prime $p$. The lemma shows that one group is p-nilpotent exactly when the other is.

A finite group is nilpotent exactly when it has a normal p-complement for every prime dividing its order. For completeness, if all those complements exist, set $L_q={\bigcap }_{p\ne q}{K}_p$, where $K_p$ is the normal p-complement. The diagonal map $G/L_q\to {\prod }_{p\ne q}G/K_p$ is injective, so $[G:L_q]$ is prime to $q$. At the same time, containment in each $K_p$ removes every prime other than $q$ from $|L_q|$. Thus $L_q$ is a normal Sylow q-subgroup. Distinct normal Sylow subgroups commute, giving the direct-product decomposition. Conversely, the direct-product decomposition of a finite nilpotent group provides all the complements. We concludeThis is the assertion that the complex group algebra detects p-nilpotency and hence nilpotency. It needs the multiplicities of the degrees, not just their distinct values.

Normal-complement consequence. If $p$ divides every nonlinear irreducible degree, the p-prime-degree characters are exactly the linear characters. ThereforeThe lemma yields a normal p-complement immediately:This proves Thompson's character-degree normal complement theorem, including the abelian case where the hypothesis on nonlinear characters is vacuous.

For an oriented three-manifold M with prescribed framed boundary endpoints P, the relative Kauffman bracket skein module is the complex vector space generated by properly embedded framed tangles, modulo framed isotopy and the local bracket relations. A crossing resolves with coefficients A and $A^{-1}$, and adjoining an unknotted circle multiplies by $\delta =-A^2-A^{-2}$. We now use the full bracket convention: the empty diagram has value one and a circle has value $\delta$. For a nonempty uncolored link this full bracket is $\delta$ times the reduced bracket used earlier. Fixing this normalization is essential for the quantum dimensions. Gluing compatible boundary pieces composes skeins; disjoint union gives tensor products. With no endpoints this is the Kauffman bracket skein module. Framing is retained, so a Reidemeister-I curl is a framing change rather than an allowed isotopy.

For $A\ne 0$, the Temperley-Lieb algebra ${\mathrm{TL}}_n(A)$ is the skein algebra of a rectangle cylinder with n marked endpoints on each horizontal side, multiplication being vertical stacking. Resolve crossings to get a basis of crossing-free matchings of its $2n$ endpoints, with every closed circle removed with factor $\delta$. There are $\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ such matchings. Conversely multiplication of these matchings by joining endpoints defines an algebra satisfying the skein relations, which proves their independence. Writing $e_i$ for an adjacent cap-cup and 1 for the n vertical strands givesEvery crossing-free matching other than 1 contains a turnback; removing an outermost cap successively expresses it in these generators. Thus these diagram rules give the usual generator presentation as well.

Put $[m]=(A^{2m}-A^{-2m})/(A^2-A^{-2})$, interpreted as the Laurent sum ${\sum }_{s=0}^{m-1}{A}^{2(m-1-2s)}$ if its denominator vanishes. Set ${\Delta }_j=(-1{)}^j[j+1]$. These quantum integers satisfy ${\Delta }_0=1$, ${\Delta }_1=\delta$, and ${\Delta }_j=\delta {\Delta }_{j-1}-{\Delta }_{j-2}$. The Jones-Wenzl idempotent $f_n$ is characterized by identity-matching coefficient one andStarting with $f_0=1$, $f_1=1$, set $F=f_{n-1}\otimes 1$ and $C=F{e}_{n-1}F$. Whenever the denominators are nonzero the recursion isHere is the induction proving its properties. Closing the last strand of $f_{n-1}$ gives $({\Delta }_{n-1}/{\Delta }_{n-2})f_{n-2}$. Consequently stacking two copies of C, and straightening the middle cap-cup, gives $C^2=\mu C$, where $\mu ={\Delta }_{n-1}/{\Delta }_{n-2}$. Also $FC=CF=C$. Expanding $(F-{\mu }^{-1}C{)}^2$ proves idempotence. The earlier turnbacks are killed by F; the last is killed because the same cap-cup calculation gives $C{e}_{n-1}=\mu F{e}_{n-1}$ and its reflected identity. Only F contributes to the identity matching, with coefficient one.

For completeness, the partial-closure formula used in that induction follows from the recursion itself. Closing the last strand of F gives $\delta f_{n-1}$; closing the last strand of C straightens its cap-cup to $f_{n-1}$. Hence partial closure of $f_n$ is $(\delta -{\Delta }_{n-2}/{\Delta }_{n-1})f_{n-1}=({\Delta }_n/{\Delta }_{n-1})f_{n-1}$. Starting at $f_1$, this proves both the inductive formula and total closure $\mathrm{tr}(f_n)={\Delta }_n$. Finally, any nonidentity matching is in the turnback ideal. If two normalized projectors kill that ideal, multiplying them gives each of them, proving uniqueness. Their normalization also gives absorption of smaller projectors on consecutive substrands. At exceptional A one must check the denominators; the recursion does not assert existence of every projector there.

A positive framing twist of an n-strand projected cable consists of n individual positive curls and $n(n-1)$ positive braid crossings. Each curl contributes $-A^3$. A crossing is $A1+A^{-1}{e}_i$, so it acts on $f_n$ as A because the turnback vanishes. This proves the depicted Jones-Wenzl twist eigenvalue identity, including its exponent:A negative framing twist acts by the inverse scalar.

Choose $A=e^{\pi i/(2r)}$, with an integer $r\ge 3$. Then $[r]=0$ but $[1],\dots ,[r-1]$ are nonzero. The projector $f_{r-1}$ exists and has trace zero. Pass to the reduced Temperley-Lieb skein theory by setting it to zero, including every diagram containing it. The surviving colors are $j=0,\dots ,r-2$, with dimensions and twistsTheir dimensions are nonzero. The recurrence decomposes $f_j\otimes 1$ into the two orthogonal channels $f_{j+1}$ and $f_{j-1}$: the first is the new projector, while its complementary cap-cup idempotent is the latter, with inclusion/projection normalized by the nonzero partial-closure scalar. At the upper endpoint the $r-1$ channel vanishes. Iterating gives a complete decomposition of every tensor power into these colors. A matching between projected colors of different sizes has a turnback on the larger side and vanishes; on the same size only the identity matching survives. Thus their endomorphism spaces are one-dimensional and all matrix blocks split. Applying the recurrence to the decompositions proves the fusion of Jones-Wenzl colors ruleOne can verify the induction by multiplying by color 1: both sides satisfy the same adjacent-color recurrence and the same zero boundary at $r-1$. Taking closures proves $d_a{d}_b={\sum }_c{N}_{ab}^c{d}_c$. Cups and caps make every color self-dual, so these fusion coefficients are symmetric in all three labels.

We next prove the Kirby color handle-slide identity, rather than assume it. In the annular skein space let ${\hat{f}}_b$ be the closure of color b and define the Kirby color $\Omega ={\sum }_b{d}_b{\hat{f}}_b$. For a fixed color a, choose inclusion/projection maps $i_c:c\to b\otimes a$, $p_c:b\otimes a\to c$ for the fusion channels, with $p_c{i}_c=1_c$ and ${\sum }_c{i}_c{p}_c=1_{b\otimes a}$. Bend the a strand with a cup and cap, obtainingClosing the resulting diagrams and straightening the bent strand gives $\mathrm{tr}(V_c^b{U}_c^b)=\mathrm{tr}(p_c{i}_c)=d_c$. Since b is simple, $V_c^b{U}_c^b=(d_c/d_b)1_b$. The symmetric fusion rule supplies all channels of $c\otimes a$, so their normalized projections are complete. ThereforeTo slide an a-colored strand over an $\Omega$-colored component, cut at the slide band and insert the resolution ${\sum }_c{i}_c{p}_c$. Carrying the band across that loop bends its two maps into U and V. In each outgoing channel c, the boxed identity replaces the sum over the old loop color b by exactly the new loop weight $d_c$; the rest of the diagrams are isotopic as framed ribbons. This proves invariance for fixed a. Summing a with its own weight proves invariance when both surgery components are colored by $\Omega$. The band framing is carried in this argument, as required for a framed handle slide.

We must also show that the stabilization normalization is nonzero. Let $H_{ij}$ be the full bracket of the zero-framed Hopf link with colors i,j. On a fusion channel k, its double braiding has eigenvalue ${\theta }_k/({\theta }_i{\theta }_j)$: the full twist of the combined ribbon equals double braiding followed by the two separate twists. Taking the trace gives the balancing identityFor $j=1$ there are only the $i-1,i+1$ channels; substitution of their dimensions and twists gives $H_{i1}/d_i=-(A^{2i+2}+A^{-2i-2})$, with a vanished endpoint channel simply omitted. A color-1 meridian therefore acts on the simple color i by that scalar. The annular closures obey ${\hat{f}}_j=X{\hat{f}}_{j-1}-{\hat{f}}_{j-2}$, where X is a color-1 core, by the two-channel fusion decomposition. Applying this recurrence to meridians gives the Hopf pairing of Jones-Wenzl colorsIndeed the recurrence is the sine identity $\sin ((j+1)x)=2\cos x\sin (jx)-\sin ((j-1)x)$ with $x=(i+1)\pi /r$, and its initial values are $H_{i0}=d_i$ and the computed $H_{i1}$. Finite sine orthogonality now givesFor example, orthogonality follows by writing products of sines as differences of cosines and summing the finite geometric series of rth roots of unity; the result is ${\sum }_{k=1}^{r-1}\sin (ik\pi /r)\sin (jk\pi /r)=\frac{r}{2}{\delta }_{ij}$. Thus no degeneracy remains in this pairing.

The Gauss sums of Jones-Wenzl colors are $p_{\pm }={\sum }_j{d}_j^2{\theta }_j^{\pm 1}$, the evaluations of the plus- and minus-one-framed $\Omega$-colored unknots. Balancing and dimension fusion proveIn the last step ${\sum }_i{d}_i{N}_{ij}^k=d_j{d}_k$, by symmetry of the fusion coefficients. In vector notation this is $HTd=p_{+}{T}^{-1}d$, with $T=\mathrm{diag}({\theta }_j)$. The matrix $H$ and vector $d$ are real, and each twist has modulus one, so conjugation gives $H{T}^{-1}d=p_{-}Td$. Apply H again and use its orthogonality to obtain $\boxed{p_{+}{p}_{-}={\mathcal{D}}^2}$. Hence neither factor vanishes.

We may use the surgery existence theorem and Kirby calculus: every closed oriented three-manifold has an integral framed-link surgery description, and two descriptions are related by handle slides and insertion/removal of disjoint plus- or minus-one-framed unknots. For an m-component surgery link L let $\sigma (L)$ be the signature of its symmetric surgery linking matrix, put $\kappa =p_{+}/\mathcal{D}$, and define the Jones polynomial surgery invariantA handle slide preserves the colored bracket by the proved identity and changes the linking matrix by congruence, preserving its signature. Adding a disjoint plus-one-framed unknot multiplies the bracket by $p_{+}$ and changes $(m,\sigma )$ to $(m+1,\sigma +1)$, so the net factor is $p_{+}/(\mathcal{D}\kappa )=1$. For a minus-one-framed unknot it is $\kappa p_{-}/\mathcal{D}=p_{+}{p}_{-}/{\mathcal{D}}^2=1$. Kirby calculus therefore proves that the displayed value is a well-defined oriented three-manifold invariant. Our normalization has ${\tau }_r(S^3)=1$; reversing orientation conjugates the value. Its construction uses precisely the bracket and colored projectors underlying the Jones polynomial.

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 $z=xy{x}^{-1}$ and $x=yz{y}^{-1}$. The third follows from these. Substituting for z gives $xyx=yxy$, henceWe use the relator $r=xyx{y}^{-1}{x}^{-1}{y}^{-1}$.

Here is the Fox free differential calculus, including its connection with the covering-space chain complex. Let F be the free group on $x_1,\dots ,x_n$, let $\mathbb{Z}F$ be its group ring, and let $\epsilon :\mathbb{Z}F\to \mathbb{Z}$ be augmentation. Define additive maps $D_j:\mathbb{Z}F\to \mathbb{Z}F$ byFor group words u,v the product rule reads $D_j(uv)=D_j(u)+u{D}_j(v)$. Differentiating $x_i{x}_i^{-1}=1$ 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 $x_i{x}_i^{-1}$ 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 isIt 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 $(D_j{r}_i{)}_{i,j}$ for any chosen presentation.

Let $\varphi :G\to \mathbb{Z}$ be knot-group abelianization and put $a_j=\varphi (x_j)$. Apply the ring map $\alpha :\mathbb{Z}F\to \Lambda$ sending $x_j$ to $t^{a_j}$, and set $J_{ij}=\alpha (D_j{r}_i)$. 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 complexTo 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 $\varphi (u)$, contributing $t^{\varphi (u)}{e}_j$. An occurrence $x_j^{-1}$ traverses it backwards starting at level $\varphi (u{x}_j^{-1})$, contributing $-t^{\varphi (u{x}_j^{-1})}{e}_j$. Their sum is exactly the evaluated Fox derivative. The fundamental identity, with $\alpha (r_i)=1$, gives $d_1{J}^T=0$.

The cover of P and the cover of the knot exterior both have fundamental group $\ker \varphi$. 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 isFor 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 $(1,0,\dots ,0)$. 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 givesThe first boundary is $(t-1,t-1)$, whose kernel is generated by $(1,-1)$. The relator boundary is $f(t)(1,-1)$, whenceFor 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 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.

Let $F$ be an oriented Seifert surface of genus $g$ for $K$, and choose a basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push a representative curve slightly in the positive normal direction to obtain $a_i^{+}$. The Seifert matrix in our row convention isThe difference $V-V^T$ is the intersection form of F, up to the convention for its orientation; in a symplectic basis it is a direct sum of unimodular two-by-two skew blocks.

Write $X=S^3\setminus \mathrm{int}N(K)$ for the knot exterior. Linking number with K gives the epimorphism ${\pi }_1(X)\to \mathbb{Z}$ sending a positive meridian to one. Let $\tilde{X}$ be the associated infinite cyclic cover, with deck transformation t. The Alexander module of a knot is ${\mathcal{A}}_K=H_1(\tilde{X};\mathbb{Z})$, regarded as a module over the Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$. Its order, the greatest common divisor of the maximal presentation minors, is the Alexander polynomial of a knot, defined up to $\pm t^k$. Equivalently these minors generate its zeroth Fitting ideal; we will obtain a square presentation, so its determinant gives the order. No assertion that $\Lambda$ is a principal ideal domain is needed.

Here is the Seifert-matrix presentation of the Alexander module with its topological justification. Cut X along F, obtaining a connected manifold Y with two boundary copies $F^{+}$ and $F^{-}$. Up to collars this is the complement of a thickening of F in $S^3$. Alexander duality gives $H_1(Y;\mathbb{Z})\cong {\mathbb{Z}}^{2g}$ and the perfect linking pairing with $H_1(F;\mathbb{Z})$. Choose generators $b_j$ satisfying $\mathrm{lk}(b_j,a_i)={\delta }_{ij}$. This homology statement does not require Y to be a handlebody. The two inclusions satisfyIndeed the first coefficients are the defining positive linking numbers; for the second, moving the two curves to opposite sides and using symmetry of linking gives $\mathrm{lk}(a_i^{-},a_j)=\mathrm{lk}(a_j^{+},a_i)$.

Stack copies of Y, identifying the positive copy of F in one layer with the negative copy in the next. This is $\tilde{X}$. The Mayer–Vietoris sequence, with all translates collected into Laurent modules, containsThe last map is injective because $\Lambda$ is an integral domain. Thus the entire Alexander module is presented by the row relations $tV-V^T$ (transpose the matrix if using column vectors). At $t=1$ its determinant is $\det (V-V^T)=1$, so the determinant is nonzero and the module is a torsion module. Consequentlywhere $\doteq$ denotes equality up to a Laurent unit. The second conclusion follows because a determinant of size $2g$ with entries linear in t has breadth at most $2g$; minimize over Seifert surfaces.

For the displayed diagram, start on the outer lower arc near the lower left undercrossing and follow the arc through the lower outer loop toward the upper right undercrossing. Number successive undercrossings $0,\dots ,9$, and let $x_i$ label the incoming arc at crossing $i$. Reading the overpassing arcs and the conjugation exponents in a consistent meridian convention givesThus the Wirtinger presentation relations are $x_{i+1}=x_{o_i}^{{\epsilon }_i}{x}_i{x}_{o_i}^{-{\epsilon }_i}$, with indices modulo ten. Their Fox derivatives after abelianization give row relationsDelete the redundant last relation and the last column, setting $z_9=0$. This is a nine-generator presentation of the Alexander module. The cofactor determinant is $t^{-6}(t^4+t^3-3t^2+t+1)$; here is a smaller elimination check. Retain $z_1,z_4$. Seven relations solve the other generators asThe two remaining rows, at undercrossings two and seven, are the matrixIts determinant is $t^{-3}(t^4+t^3-3t^2+t+1)$, differing from the original cofactor only by a Laurent unit. ConsequentlyThe factor has value one at $t=1$ and is reciprocal, as required. Its Conway-normalized Alexander polynomial is $t^2+t-3+t^{-1}+t^{-2}$, and its knot determinant is three.

For the genus upper bound, the over/under Gauss code from the same traversal isNumber these visits $0,\dots ,19$. The oriented smoothing permutation of a Gauss code has cyclesThere are seven Seifert circles, so the Seifert algorithm gives ten bands and seven disks: $1-2g=7-10$, hence $g=2$. The Alexander breadth bound on Seifert genus supplies the matching lower bound from breadth four. Therefore $\boxed{g_s(K)=2}$.

Modulo two, the primitive Alexander polynomial becomes $t^4+t^3+t^2+t+1$. It has neither zero nor one as a root, and it is not $(t^2+t+1{)}^2=t^4+t^2+1$, the only possible product of irreducible monic quadratics over ${\mathbb{F}}_2$. It is therefore irreducible over ${\mathbb{F}}_2$, hence over $\mathbb{Q}$ and $\mathbb{Z}$ by Gauss lemma for polynomials. The full-degree irreducible Alexander polynomial implies a prime knot criterion now applies, since its breadth is exactly twice the genus. Thus the knot is prime. The full-degree condition is needed to rule out a nontrivial summand with unit Alexander polynomial.

To test reflection, calculate the Jones polynomial from the same crossing data. The diagram has nine positive crossings and one negative crossing, so its writhe of a link diagram is eight. Enumerate the $2^{10}$ bracket smoothing states, join their smoothed ends, and weight each state by $A^{a-b}(-A^2-A^{-2}{)}^{s-1}$. Correct by $(-A^3{)}^{-8}$ and put $t=A^{-4}$. This givesAs checks, $V_K(1)=1$ and $V_K(-1)=-3={\Delta }_K(-1)$. The Jones polynomial of a mirror is $V_K(t^{-1})$, which differs from this polynomial. Therefore the knot is not equivalent to its reflection. Reversing all crossing conventions reciprocates the displayed Jones polynomial and leaves the conclusion unchanged.

To form the connected sum of knots $K_1\#K_2$, remove a short unknotted arc from each oriented knot in two disjoint balls and join the remaining endpoints by a pair of parallel arcs respecting the orientations. Equivalently, a sphere meeting a knot in two points cuts it into two one-string tangles; close each by an arc in the sphere to recover its summands. Sliding the small joining balls along the knots shows that the isotopy type does not depend on these choices. The unknot is the identity for this operation. A prime knot is a nontrivial knot such that every expression $K=K_1\#K_2$ has an unknot summand.

The Seifert algorithm supplies a Seifert surface for every oriented knot: smooth the crossings in their oriented directions, span the resulting Seifert circles by disks at different heights, and restore each crossing with a half-twisted band. If needed, join components by tubes away from the boundary. Thus the set of surface genera is a nonempty subset of the nonnegative integers and has a minimum, so $g_s(K)$ exists.

We first prove additivity of Seifert genus, which also supplies a terminating measure for decomposition. Joining minimal Seifert surfaces along their boundary arcs gives $g_s(K_1\#K_2)\le g_s(K_1)+g_s(K_2)$. For the reverse inequality, let $F$ be a minimal-genus Seifert surface for the sum and $S$ a splitting sphere of a knot. Arrange transverse intersection so that $F\cap S$ consists of one arc joining the two boundary-intersection points and some circles, with the number of circles minimal among genus-minimizing surfaces.

Each circle separates the sphere into two disks; since it is disjoint from the intersection arc, one disk contains neither marked endpoint. Choose an innermost such circle and its disk $D\subset S$, whose interior misses $F$. If its boundary were essential in $F$, compressing $F$ along $D$ would lower the genus of its component with boundary K: for a nonseparating curve genus drops by one, and for a separating essential curve the removed boundaryless side has positive genus. This contradicts minimal genus. Its boundary therefore bounds a disk in $F$. Replace that disk by a small push-off of $D$, removing the intersection circle without changing genus or boundary, again contradicting minimal intersection. Thus no circles remain.

Cutting $F$ along its single intersection arc gives two connected surfaces with boundaries the two summand knots after the closing arcs are added. Euler characteristic shows that their genera add to $g(F)$. Each genus is at least the minimum genus of its summand, soGenus zero means that the knot bounds a disk and is the unknot. Every nontrivial knot therefore has positive genus. Induct on this nonnegative integer: a nontrivial nonprime knot splits into two nontrivial summands, each with strictly smaller genus by the equality above, and induction decomposes both into primes. This proves every knot is a finite connected sum of prime knots, with the unknot the empty sum.

The same induction does not follow from the three suggested replacements. For crossing number of a knot, joining minimal diagrams gives only subadditivity; a reverse inequality giving strict descent for every nontrivial summand is not provided by this argument. The Jones polynomial of a connected sum does make its breadth of a Laurent polynomial additive, but the needed implication that zero breadth forces the unknot is the Jones unknot detection problem. For the Alexander polynomial, even this implication is false: nontrivial untwisted Whitehead doubles have Alexander polynomial one and hence breadth zero. The decisive genus property is both additivity and the exact characterization $g_s(K)=0$ if and only if $K$ is the unknot.

For uniqueness use complete decomposing sphere systems for a knot. Cut each system and cap the marked boundary spheres with straight arcs; the resulting oriented factors are prime knots. Compare two systems in transverse position, and use the prime-ball exchange lemma to simplify their intersection. Choose an innermost intersection circle bounding a disk in a sphere of one system and lying in a prime-factor ball of the other. If the disk misses the knotted arc it cuts off an empty ball. If it meets the arc once it cuts the prime factor into two knots, so one is an unknot. Replace the old ball by the side carrying the same prime factor, perturb its boundary slightly, and remove that intersection circle. This preserves the unordered factors: the discarded empty or trivial ball merely changes the joining arc, and any intervening capped boundaries still carry their original factors. If the chosen disk contains both marked points, take another innermost disk without marked points, or the complementary disk when the intersection circle is the only one.

Iteration yields an innermost prime-factor ball disjoint from both systems. Its arc has no nontrivial further decomposition, so it represents one factor in each. The boundary can be included in both systems using the same ball exchange; replace its arc by a straight boundary arc and remove this common factor. Induction on the number of factors matches the remaining factors. When the systems already have no intersections, start directly with an innermost ball and the same removal step. All replacements preserve the arc orientation, so the match is of oriented knots, not only unoriented types. Hence the prime decomposition of knots is unique up to reordering and oriented link isotopy.