Residual Torsion-Free Nilpotence, Bi-Orderability and Pretzel Knots

The residual torsion-free nilpotence of the commutator subgroup of a knot group has played a key role in studying the bi-orderability of knot groups. A technique developed by Mayland provides a sufficient condition for the commutator subgroup of a knot group to be residually-torsion-free nilpotent using work of Baumslag. In this paper, we apply Mayland's technique to several genus one pretzel knots and a family of pretzel knots with arbitrarily high genus. As a result, we obtain a large number of new examples of knots with bi-orderable knot groups. These are the first examples of bi-orderable knot groups for knots which are not fibered or alternating.


Introduction
Let J be a knot in S 3 .The knot exterior of J is M J := S 3 − ν(J) where ν(J) is the interior of a tubular neighborhood of J, and the knot group of J is π 1 (M J ). Denote the Alexander polynomial of J by ∆ J .
A group Γ is nilpotent if it's lower central series terminates (is trivial) after finitely many steps.In other words, for some non-negative integer n, where Γ 0 = Γ and Γ i+1 = [Γ i , Γ] for each i = 0, . . .n − 1.A group Γ is residually torsion-free nilpotent if for every nontrivial element x ∈ Γ, there is a normal subgroup N Γ such that x / ∈ N and G/N is a torsion-free nilpotent group.This paper concerned with when the commutator subgroup of a knot's group is residually torsion-free nilpotent, which has applications to ribbon concordance [15] and the bi-orderability of the knot's group [26].
Several knots are known to have groups with residually torsion-free nilpotent commutator subgroups.The commutator subgroup of fibered knot groups are finitely generated free groups, which are residually torsion-free nilpotent [28].Work of Mayland and Murasugi [30] shows that the knot groups of pseudo-alternating knots, whose Alexander polynomials have a prime power leading coefficient, have residually torsion-free nilpotent commutator subgroups; pseudo-alternating knots are defined in Section 3. The knot groups of two-bridge knots have residually torsion-free nilpotent commutator subgroups [20].
There is also the following obstruction to a knot's group having residually torsionfree nilpotent commutator subgroup.Proposition 1.1.If J is a knot in S 3 with trivial Alexander polynomial, then the commutator subgroup of π 1 (M J ) cannot be residually torsion-free nilpotent.
Proof.Let G be the commutator subgroup of π 1 (M J ).Let M ∞ be the infinite cyclic cover of M J , the covering space of M J corresponding to G so that π 1 (M ∞ ) = G; see [36,Chapter 7] for details.

Since the Alexander polynomial of
It follows that every term of the lower central series of G is isomorphic to G. Suppose N G is a proper normal subgroup of G.For each term of the lower central series of G/N , (G/N ) i ∼ = G i /N ∼ = G/N = 1 so G/N cannot be nilpotent.Thus, G is not residually torsion-free nilpotent.
Given the integers k 1 , k 2 , . . ., k n , define P (k 1 , k 2 , . . ., k n ) to be the pretzel knot represented in the diagram in Figure 1.Mayland [29] describes a technique to examine the commutator subgroup of the group of a knot bounding an unknotted minimal genus Seifert surface; see section 2. In fact, this is the technique Mayland and Murasugi used to prove their result for pseudo-alternating knots [30].Applying Seifert's algorithm to the diagram in Figure 1 yields an unknotted minimal genus Seifert surface [12] making pretzel knots ideal candidates for Mayland's technique.
1.1.Possible Generalizations.The techniques used here have a few limitations.First, while our method can be applied to many families of genus one pretzel knots on a case by case basis, this method does not lend itself well to generalizing to all genus one pretzel knots since many of the details depend on the arithmetic properties of p, q and r.Secondly, Mayland's method requires a couple conditions (an unknotted Seifert surface satisfying the free factor property and an Alexander polynomial with prime power leading coefficient) which may not be necessary for a knot group to have residually torsion-free nilpotent commutator subgroup.Nevertheless, we make the following prediction for genus one pretzel knots.
Conjecture 1.5.If J is a genus one pretzel knot then the commutator subgroup of π 1 (M J ) is residually torsion-free nilpotent if and only if the Alexander polynomial of J is nontrivial.

Application to Bi-Orderability.
A group is said to be bi-orderable if there exists a total order of the group's elements invariant under both left and right multiplication.The following fact [7,Theorem B] follows from work of Linnell, Rhemtulla, and Rolfsen [26] and has been instrumental in determining the bi-orderability of several knot groups [35,8,20].
Theorem 1.6.[7, Theorem B] Let J be a knot in S 3 .If π 1 (M J ) has residually torsion-free nilpotent commutator subgroup and all the roots of ∆ J are real and positive then π 1 (M J ) is bi-orderable.
Furthermore, Ito obtained the following obstruction to a knot group being biorderable when the knot is rationally homologically fibered [18]; see section 2 for definition of rationally homologically fibered.
Theorem 1.7.[18, Theorem 2] Let J be a rationally homologically fibered knot.If π 1 (M J ) is bi-orderable then ∆ J has at least one real positive root.
∆ J has two positive real roots when N < 0 and two non-real roots when N > 0.
Applying Proposition 1.8 to the results in Theorem 1.2 yields the following corollary.
We also have the following corollary to Theorem 1.4.
Details of the proof of Corollary 1.10 are provided in section 4.

1.3.
A Possible Connection of Bi-Orderability to Branched Covers.Given a knot J in S 3 , let Σ n (J) be the n-fold cyclic cover of S 3 branched over J; see [36,Chapter 10] for the definition and construction of a cyclic branched cover.Part of the motivation for studying the bi-orderability of pretzel knots is to investigate the following questions.
Question 1.11 is resolved here.
Remark 1.14.Question 1.11 is still unanswered for fibered knots and alternating knots.
Question 1.12 remains unresolved as of the writing of this paper.However, some important remarks can be made about this question.
Proposition 1.15.Suppose J is the P (2p + 1, 2q + 1, 2r + 1) pretzel knot with p < −1 and 1 ≤ q ≤ r. π 1 (Σ 2 (J)) is left-orderable if and only if −p ≤ q.Thus, if p < −1 and the double branched cover of J does not have left-orderable fundamental group, then q < −p so N as defined in (1.1) is negative.Therefore, if Conjecture 1.5 is true, π 1 (M J ) would be bi-orderable when q < −p by Proposition 1.8.In particular, if Conjecture 1.5 is true, it's not likely that any non-alternating genus one pretzel knot would be a counterexamples for Question 1.12.
There is some evidence that genus one pretzel knots with no left-orderable cyclic branched covers do exists.It is conjectured [3] that given a prime orientable closed rational homology sphere Y , π 1 (Y ) is not left-orderable if and only if Y is an Lspace, and Issa and Turner show that the cyclic branched covers of the P (−3, 3, 2r+ 1) pretzel knots are all L-spaces [17].
1.4.Outline.In section 2, we review how Mayland's technique [29] can be used to analyze when the commutator subgroup of a knot group is residually torsion-free nilpotent.In section 3, we apply this technique to genus one pretzel knots and prove Theorem 1.2 and Theorem 1.13.In section 4, we prove Theorem 1.4.Appendix A contains the proofs of some key lemmas.We also provide a chart of our results in appendix B. 1.5.Acknowledgments.The author would like to thank Cameron Gordon for his guidance and encouragement throughout this project.The author would like to thank Ahmad Issa and Hannah Turner for many helpful conversations.The author would like to thank the anonymous Algebraic & Geometric Topology referee for helpful comments and critiques.This research was supported in part by NSF grant DMS-1937215.

Preliminaries on Mayland's Technique
Mayland used a description of the commutator subgroup of a knot group to investigate when they are residual finite [29].In this section, we show how Mayland's technique can be used to find a sufficient condition for the commutator subgroup of a knot group to be residually torsion-free nilpotent.
2.1.Mayland's Technique.Let J be a knot in S 3 , and suppose J bounds a minimal genus Seifert surface S such that S is unknotted, in other words, π 1 (S 3 \S) is a free group.Let Ŝ = M J ∩ S. Let G be the commutator subgroup of π 1 (M J ).
Let U be the image of a bi-collar embedding Ŝ × [−1, 1] → M J where Ŝ is the image of Ŝ × {0}, and let M S = M J \ Ŝ. Denote images of Ŝ × (0, 1] and Ŝ × [−1, 0) in M S as U + and U − respectively.Let X = π 1 (M S ) which is a free group of rank 2g where g is the genus of J. Consider the inclusion maps i + : For each integer n, let X n be a copy of X, H n ⊂ X n be a copy of H, and K n ⊂ X n be a copy of K.The fundamental groups of U , U + and U − are canonically isomorphic, and since S has minimal genus, i + * and i − * are injective.Therefore, H n and K n+1 are identified with a rank 2g free group F .By a result of Brown and Crowell [5, Theorem 2.1], G is an amalgamated free product of the following form. (2.1) Baumslag provides the following sufficient condition [2, Proposition 2.1(i)] for a group to be residually torsion-free nilpotent when G is an ascending chain of parafree subgroups; see [1,2] for definition and discussion of parafree groups.
Proposition 2.1.[2, Proposition 2.1(i)] Suppose G is a group which is the union of an ascending chain of groups as follows.
Suppose each G n is parafree of the same rank.If for each non-negative integer n, For each non-negative integer m, define Z m as follows. (2.2) The direct limit of the Z m 's is isomorphic to G. Furthermore, since i + * and i − * are injective, the natural inclusion Z m → Z m+1 is an embedding so G is an ascending chain of subgroups as follows.
It immediately follows that A is a free factor of B if and only if every (equivalently, at least one) free basis of A extends to a free basis of B. A theorem of Mayland [29,Theorem 3.2] provides the following sufficient conditions for each Z m to be parafree.
for some prime p and non-negative integer l, then for every non-negative m, Z m is parafree of rank 2g.
The knot J is rationally homologically fibered if the induced map on homology, , is an isomorphism.Let S + be a Seifert matrix representing i + h so that S − := S T + is a Seifert matrix representing i − h .S + is also a presentation matrix for the abelian group Denote the standard form of the Alexander polynomial of J by ∆ J .For some non-negative integer k, [33,Chapter 6]).
Proposition 2.3.Suppose J is a knot in S 3 .The following statements are equivalent: (a) J is rationally homologically fibered.Proposition 2.4.When J is rationally homologically fibered, Proof.When J is rationally homologically fibered, The Seifert surface S is said to satisfy the free factor property if H and K are free factors of H[X, X] and K[X, X] respectively.Note that this property is independent of the orientation of S. A sufficient condition for the residual torsion-free nilpotence of G can be summarized as follows.
Proposition 2.5.Suppose J is a rationally homologically fibered knot in S 3 with unknotted minimum genus Seifert surface S. If S satisfies the free factor property and |∆ J (0)| is a prime power, then the commutator subgroup, G, is residually torsion-free nilpotent.
Proof.Suppose J is a rationally homologically fibered with unknotted minimum genus Seifert surface S satisfying the free factor property, and suppose |∆ J (0)| is a prime power.
Define Z m for each non-negative integer m as in (2.2).By Proposition 2.4, |X : H[X, X]| and |K : H[X, X]| are prime powers since J is rationally homologically fibered.Thus, by Proposition 2.2, each Z m is parafree of rank twice the genus of J.
By (2.4), 2.2.Pseudo-Alternating Knots.A special alternating diagram is an alternating link diagram in which all of the crossings have the same sign.Any link with such a diagram is called a special alternating link.The Seifert surface described by performing Seifert's algorithm on special alternating diagram is a primitive flat surface.A generalized flat surface is any surface which can be obtained by combining some number of primitive flat surfaces by Murasugi sums.A link which bounds a generalized flat surface is a pseudo-alternating link.Alternating links are pseudoalternating links.However, all torus links, many of which are not alternating, are also pseudo-alternating links.
Pseudo-alternating knots are rationally homologically fibered and bound surfaces satisfying the free factor condition [30,Theorem 2.5].Therefore, the knot group of a pseudo-alternating knot, whose Alexander polynomial has a prime power leading coefficient, has residually torsion-free nilpotent commutator subgroup.

Genus One Pretzel Knots
Let J be the P (2p + 1, 2q + 1, 2r + 1) pretzel knot for some integers p, q, and r with 1 ≤ q ≤ r and p = −1 or 0. Let S be the unknotted genus 1 surface depicted in Figure 2, which we refer to as the standard Seifert surface of J.For the genus one pretzel knots which aren't two-bridge knots, the standard Seifert surface is the unique Seifert surface of minimal genus up to isotopy and [13].
By Gabai [11], S must be a minimal genus Seifert surface so χ(S) = −1.Analyzing the effect of a Murasugi sum on the Euler characteristic yields Since S 1 and S 2 are not disks, neither S 1 or S 2 has positive Euler characteristic.It follows that χ(S 1 ) = χ(S 2 ) = 0 so S 1 and S 2 are both annuli.
The boundary of a Murasugi sum of two annuli is a double twist knot which is alternating.Thus, P (2p + 1, 2q + 1, 2r + 1) is alternating, which is a contradiction.
Since J is pseudo-alternating when p ≥ 0, we will only need to focus on the case when p is negative.
3.1.Mayland's Technique for Genus One Pretzel Knots.Define M J , M S , X, H and K as in section 2.Here we offer a concrete description of the maps on fundamental groups i + * and i − * for genus one pretzel knots.This is the same discription used by Crowell and Trotter in [9].Choose a base point z on the lower part of S, and let x and y be the classes generating π 1 (S, z) represented by the loops indicated in Figure 2. Let z + and z − be push-offs of z of each side of S. Let z be the base point of M S obtained by shifting z tangentially along S through ∂S.Let δ + and δ − be arcs connecting z to z + and z − respectively; see Figure 3. Finally, let a and b be the indicated classes generating π 1 (M S , z ).
By slightly isotoping elements of π 1 (S, z) off of S, π 1 (U + , z + ) and π 1 (U − , z − ) are canonically identified to π 1 (S, z) which is a rank two free group, F , generated by x and y.The group X := π 1 (M S , z ) is a rank two free group generated by a and b.
Let N = det S + = det S − .Up to multiplication by a signed power of t, the Alexander polynomial of J is When N = 0, J is rationally homologically fibered by Proposition 2.3.Simply considering the integer N can provide useful information.Proposition 3.2.When N = 0, G is not residually torsion-free nilpotent.
Proof.∆ J (t) = 1 when N = 0 so G cannot be residually nilpotent by Proposition 1.1.Proof.Let S be the standard Seifert surface of J, and define X, H, and K as in section 2. Each of these are rank 2 free groups.Suppose S satisfies the free factor property. When Since H is a free factor of H[X, X] and both are rank 2 free groups, H = H[X, X] = X.Similarly, since K is a free factor of K[X, X] and both are rank 2 free groups, K = X.This implies that i + * and i − * are isomorphisms.Thus, π 1 (M J ) is an extension Z described by the following short exact sequence.
The Stallings fibration theorem [37] implies that J is a genus 1 fibered knot [37].However, the only genus 1 fibered knots are the trefoil and the figure 8 knot [6,14] which is a contradiction since we are assuming J is not a two-bridge knot.
In light of Proposition 2.5, to prove the commutator subgroup of π 1 (M J ) is residually torsion-free nilpotent, it is sufficient to show S satisfies the free factor property.

Outline of Procedure.
In each case, we use the same basic procedure, outlined below, to analyze whether or not S satisfies the free factor property.
(1) Find a presentation matrix for X/H[X, X] of the form u v 0 w or u 0 v w using row operations.Note, u and w can always be made positive.Thus, X/H[X, X] is isomorphic to (Z/uZ) × (Z/wZ).The Z/uZ factor is generated by the class of a in X/H[X, X], and the Z/wZ factor is generated by the class of b.
(2) Since X/H[X, X] is abelian, the set C, defined below, is a set of coset representatives of H[X, X].
Given x ∈ X, denote by x, the coset representative of x in C. Define where c ∈ C and x ∈ {a, b}.From this, we find the following free basis for H[X, X] using the Reidemeister-Schreier method; see [21] for details.
(3) Use the Reidemeister-Schreier rewriting process to rewrite the generating set of H from (3.1).A word α ∈ H, where α = α s1 1 • • • α s k k with α i ∈ {a, b} and s i = ±1, can be rewritten as where (4) Determine if the generating set of H can be extended to a free basis of H[X, X]. (5) Repeat this procedure for K.
When the free bases of H and K can be extended to free bases of H[X, X] and K[X, X] respectively, S satisfies the free factor property.If the chosen basis of either H or K fails to extend then S cannot satisfy the free factor property.
Proof.Suppose J is P (−5, 7, 7).From (3.1), we have that The abelian group X/H[X, X] has presentation matrix Therefore, the set {α H , β H , x 0 , x 1 , x 3 , x 5 } is a generating set of six elements for H[X, X], and thus, is a free basis.It follows that After row reductions, X/K[X, X] has presentation matrix 1 −3 0 5 .
From this we get a free basis of K[X, X] as follows. x for k = 5 Rewriting α K and β K , we get . Therefore, S satisfies the free factor property.
Suppose J is P (−5, 7, 9).X/H[X, X] has presentation matrix which becomes 1 0 0 4 after row operations.By Reidemeister-Schreier, we obtain the free basis {x 0 , x 2 , x 3 , x 4 } where x i = b i ab −i for i = 0, . . ., 3 and A similar argument shows K is a free factor of K[X, X].Therefore, S satisfies the free factor property.Lemma 3.5.If J is a P (−3, 3, 2r + 1) pretzel knot then S satisfies the free factor property.
Proof.From (3.1), we have that The abelian group X/H[X, X] has presentation matrix 1 0 0 2 when r is even and 1 −1 0 2 when r is odd.
Using C = {1, b} as a set of coset representatives, we apply Reidemeister-Schreier to obtain a free basis of H[X, X], B = {x 0 , x 1 , x 2 } When r is even, When r is odd, In either case, the set {α X/K[X, X] has presentation matrix 2 0 0 1 .
Using C = {1, a} as a set of coset representatives, we get the free basis of K[X, X], B = {x 0 , x 1 , x 2 } where Thus, Therefore, S satisfies the free factor property.
Then, S satisfies the free factor property.
To simplify computations, we modify this basis by multiplying some of the elements by b −C on the right, and obtain a free basis of H[X, X], B = {x 0 , . . ., x C } where We can rewrite α H and β H as Since q ≥ 4, the generator x 0 appears once in the expression for α H and does not appear in the expression for β H . Also, since q − 2 < C − 1 and qk − 1 < C − 1 for all k = 1, . . ., q − 2, x C−1 only appears once in the expression for β H . Thus, After row reductions, X/K[X, X] has presentation matrix We can rewrite α K and β K as The generator x q appears once in the expression for α K and does not appear in the expression for β K .Also, x C only appears once in the expression for β K .
Therefore, the set {α K , β K , x 0 , . . ., x q−1 , x q+1 , . . ., x C−1 } is a free basis of K[X, X] so K is a free factor of K[X, X].Therefore, S satisfies the free factor property.
Proof of Theorem 1.13.By Lemma 3.9, J q has a Seifert surface satisfying the free factor property.The Alexander polynomial of J q is N t 2 + (1 − 2N )t + N where N = −(q − 1) 2 so J q is rationally homologically fibered and ∆ Jq has two positive real roots.
We have the free basis . The abelianization of E is Z, but E is not abelian since x 1 and x 2 do not commute.Thus, E is not free, and Γ isn't free either, which is a contradiction.
Therefore, H is not a free factor of H[X, X], and S does not satisfy the free factor property.
When J is P (−5, 7, 25), H[X, X] has a free basis x 0 = a, x 1 = bab −1 , x 2 = b 2 ab −2 , x 3 = b 3 ab −3 , and x 4 = b 4 .Under this basis {α H , β H } can be extended to a free basis of H[X, X] if and only if {α H , β H α H } can be extended to a free basis.However, an argument similar to the previous case shows that β H α H cannot be extended to a basis of H[X, X].
Therefore, H is not a free factor of H[X, X], and S does not satisfy the free factor property.
When J is P (−5, 7, 13) or P (−5, 7, 21), H[X, X] has free basis x 0 = a, x 1 = bab −1 , and x 2 = b 2 .Thus, the set {x 0 , x 1 , x −1 2 x 1 x 0 } is also a free basis of H[X, X].Denote x −1 2 x 1 x 0 by y.Using the basis {x 0 , x 1 , y}, 0 .An argument similar to the previous cases shows that α H cannot be extended to a basis of H[X, X].Therefore, H is not a free factor of H[X, X], and S does not satisfy the free factor property.
The proofs of the other cases are similar to the cases above.Here we provide the elements obstructing the free factor property.
Proof.For each of these knots, N = 0 so this follows from Proposition 3.2.

Higher Genus Pretzel Knots
In this section, we prove Theorem 1.4 which presents a family of pretzel knots with arbitrarily high genus whose groups have residually torsion-free nilpotent commutator subgroups.
Let k be a positive integer, and let r be any integer.Suppose J is the 2k + 1 parameter pretzel knot P (3, −3, . . ., 3, −3, 2r + 1) with genus k Seifert surface S as shown in Figure 4. Define X, H and K as in section 2.
Proof of Theorem 1.4.X is a free group of rank 2k with generating set {a 1 , . . ., a 2k } as show in Figure 4.By choosing a suitable free basis for π 1 (S), the subgroup H where the jth Z/2Z factor is generated by the class of a 2j in X/H[X, X], and when i is odd, the class of a i is trivial.Define The following set is a set of coset representatives of H[X, X].
From C, we find a free basis B of elements of the form x k,σ := a σ a k a σ a k −1 .We point out a few important examples of basis elements.For i odd, For i even, For i odd and j even, Using the basis B rewrite the α i as which can be extended to the free basis B of H[X, X] defined below. where A similar argument shows K is a free factor of K[X, X].Thus, S satisfies the free factor property.
From (4.1), we compute |X : H[X, X]| = 2 k so by Proposition 2.3, J is rationally homologically fibered.Thus, S is an unknotted minimal genus Seifert surface, and J is rationally homologically fibered.It follows from Proposition 2.5 that the commutator subgroup of J is residually torsion-free nilpotent.
Proof of Corollary 1.10.From the Seifert matrix (4.1), we compute the following Alexander polynomial.
It follows from Theorem 1.4 and Theorem 1.6 that π 1 (M J ) is bi-orderable.
Proof.From (3.1), we have that The abelian group X/H[X, X] has a presentation matrix Using the rewriting process, we have that and (Note that since p < −2, C > r + 2 so x r+2 is defined.)We can extend {α H , β H } to the set {α H , β H , x 1 , . . ., x C−2 , x C } which is a free basis of H[X, X] so H is a free factor of H[X, X].
X/K[X, X] has a presentation matrix .
Let l = −p − 2 so C = l(r + 2) + 2. Note that l is a positive integer.We obtain a free basis of K[X, X].Using the rewriting process, we have that and β K = x 0 x −1 C x l(r+1)+2 x lr+2 x l(r−1)+2 • • • x 2l+2 x l+1 .The set {α K , β K , x 1 , . . ., x l , x l+2 , . . ., x C } is a free basis of K[X, X] so K is a free factor of K[X, X].Thus, S satisfies the free factor property.which is a contradiction.
Thus x (0) = x 2 only appears once in β H so the set {β H α H , β H , x 1 , x 3 , . . ., x N } is a free basis of H[X, X].Therefore, H is a free factor of H[X, X].
Proof.Let c be the integer such that q = 3c.X/K[X, X] has a presentation matrix so ζ(i) is distinct for each i = 0, . . ., 3c − 2. Thus x ζ(0) = x 1 only appears once in β K so the set {β K α K , β K , x 2 , . . ., x N } is a free basis of K[X, X].Therefore, K is a free factor of K[X, X].
Proof.Let c be the integer such that q = 3c + 1. X/K[X, X] has a presentation where so η(i) is distinct for each i = 0, . . ., 3c − 2. Thus x η(0) = x 2 only appears once in β K so the set {β K α K , β K , x 1 , x 3 , . . ., x N } is a free basis of K[X, X].Therefore, K is a free factor of K[X, X].
nilpotent commutator subgroup.If in addition, N < 0, then the knot's group is bi-orderable.

Figure 1 .
Figure 1.A Pretzel Knot Diagram: The integers in the boxes indicate number of right-hand half-twist when positive and lefthand half-twist when negative.

Proposition 2 . 2 .
[29, Theorem 3.2]  If H and K are free factors of H[X, X] and K[X, X] respectively, and if additionally,

Proof.
The equivalence of (b) and (c) follows from (2.3).Since S + is a presentation matrix for X/H[X, X], |X : H[X, X]| is finite if and only if | det(S + )| = 0.It follows that (a) and (b) are equivalent.Since d 2g = d 0 = det(S + ), deg ∆ J = 2g if and only if det(S + ) = 0 so (a) and (d) are equivalent.

Proposition 3 . 3 .
If |N | = 1, then the Standard Seifert surface S does not satisfy the free factor property.
. From this, we get C = {1, b, b 2 , b 3 , b 4 } as a set of coset representatives of H[X, X].We apply Reidemeister-Schreier to obtain the following free basis of H[X, X].B = {ab, ba, b 2 ab −1 , b 3 ab −2 , b 4 ab −3 , b 5 } Label the basis elements as follows: x k := b k ab 1−k for 0 ≤ k ≤ 4 and x 5 := b 5 .Now, we can rewrite α H and β H in terms of B.

B
= {x 0 , . . ., x C } where x k := a k ba l−k when 0 ≤ k ≤ C − 1 and x C := a C .
the Seifert matrices for i +