HIGMAN’S EMBEDDING THEOREM AND DECISION

PROBLEMS

ALEX BURKA

Abstract. This paper exposits Higman’s Embedding Theorem, which estab-

lishes a correspondence between finitely generated subgroups of finitely pre-sented groups and recursively presented groups. To that end we present the

basic constructions in combinatorial group theory and the correspondence be-

tween Turing machines and their associated semigroups, ultimately combiningthe concepts to prove Higman’s theorem. Our discussion culminates in a host

of applications to decision problems in group theory and topology.

Contents

1. Introduction 12. AFPs, HNN Extensions, and Normal Forms 32.1. Amalgamated Free Products 32.2. HNN Extensions 63. Turing Machines and Associated Semigroups 104. Higman’s Embedding Theorem 125. Applications to Decision Problems 195.1. The Word Problem 195.2. Markov Properties and a Universal F.P. Group 195.3. The Homeomorphism Problem Revisited 21Acknowledgements 22References 22

1. Introduction

Our story begins not with computability theory, nor does it begin with grouptheory per se. Rather, it begins with Max Dehn’s 1907 paper Analysis situs, whichdetailed the first rigorous proof of the Classification of Surfaces. Such a neat char-acterization of all imaginable closed, connected 2-manifolds raises an interest inthe same problem—the so-called homeomorphism problem—for higher dimensions.Can we classify (closed, connected) 3-manifolds? What about when n ≥ 4?

Dehn began investigating the homeomorphism problem for 3-manifolds in his1910 paper On the Topology of Three-Dimensional Spaces. Though Dehn did notsuccessfully classify 3-manifolds,1 he introduced some far-reaching techniques thatare still relevant in geometric topology. The most important of these techniques wasthe Dehn surgery; indeed, all closed, connected, orientable 3-manifolds result from

1This result arrived some fifty years later, due to the independent efforts of W. B. R. Lickorish

and Andrew H. Wallace.

1

2 ALEX BURKA

some Dehn-surgery on the 3-sphere. Performing a Dehn surgery on a 3-manifoldM involves removing a tubular neighborhood of a link L and then attaching theboundary of a solid torus to each remaining component of M \ L.

The Classification of Surfaces and Dehn surgeries share a common group-theoreticapparatus. The first result tells us all surface groups are finitely presented groupswith one relator. The second technique bears a close relationship with Dehn’stheory of tame knots, whose knot groups are given by Wirtinger presentations—in particular, by finite presentations. Indeed, Dehn recognized finitely presentedgroups as a salient feature of interesting topological questions and thus called foran investigation of them in their own right in his 1911 paper On Infinite Discon-tinuous Groups. He presented three purely group-theoretic problems—the WordProblem, Conjugacy Problem, and Isomorphism Problem—to define the course offuture research. We are interested in the first:

Problem 1.1. (The Word Problem) “An element of the group is given as a productof generators. One is required to give a method whereby it may be decided in a finitenumber of steps whether this element is the identity or not” [5].

Certain groups and classes of groups were shown to have solvable word problem,such as trefoil knot groups, braid groups, and 1-relator groups, among others. Ofcourse, what we mean by a solvable word problem demands no further attention;one simply produces a procedure distinguishing an unreduced word as the identity.But Dehn himself acknowledged the immense undertaking of solving the word prob-lem for all finitely presented groups. He once remarked, “Solving the word problemfor all groups may be as impossible as solving all mathematical problems,” chal-lenging the decidability component of Hilbert’s Program to come some ten yearslater. Turing’s exhibition of an uncomputable problem—the Halting Problem—inhis groundbreaking 1936 paper already crippled this facet of Hilbert’s program.If we accept the Church–Turing thesis, Turing’s work formally establishes what itmeans for a problem to be algorithmically unsolvable. The question remains: couldthere exist a finitely presented group with a word problem no algorithm can solve?

As it turns out, there do exist finitely presented groups which have unsolvableword problem. This result, known as the Novikove–Boone Theorem, is in part thecentral result of this paper. But the Novikove–Boone Theorem only scratches thesurface of a deep correspondence between computability theory and algebra ex-pressed in Higman’s Embedding Theorem, which asserts that recursively presentedgroups are exactly the finitely generated groups that embed into a finitely presentedgroup. This result suffices to prove the Novikove–Boone Theorem, from which ahost of other undecidability results in group theory and topology follow.

In the discussion to follow, we will present a proof of this remarkable fact. Beforereaching this result, we first exposit the essentials of combinatorial group theory.In particular, we define amalgamated free products and HNN extensions, and theninvestigate their normal form properties. A brief exposition of computability the-ory follows, in which we emphasize the natural correspondence between a Turingmachine and its associated semigroup. The theory developed in the first sectionbecomes relevant when we wish to extend this correspondence between Turing ma-chines and their associated semigroups, to associated semigroups and a particularseries of amalgamated free products and HNN extensions. These correspondencesform the foundation of the proof of sufficiency in Higman’s theorem. In the finalsections, we consider applications to decision problems.

HIGMAN’S EMBEDDING THEOREM AND DECISION PROBLEMS 3

This paper requires no background in combinatorial group theory other thana working understanding of free groups and group presentations. Albeit self-contained, our section on computability does not motivate the definitions therein.An intuitive grasp of the concepts in [10] are thus essential. Finally, we assumesome topology in Subsection 5.3, but not all details are essential in understandingthe significance of Theorem 5.3 in light of our earlier comments.

2. AFPs, HNN Extensions, and Normal Forms

Before introducing amalgamated free products and HNN extensions, we reviewsome basic concepts from combinatorial group theory. Given a set X, the freegroup F satisfies the universal property that every set map X to a group G factorsthrough a unique homomorphism F → G; denote this group 〈X〉. Given sets X,Y ,we have 〈X〉 ∼= 〈Y 〉 iff |X| = |Y |, so the rank of a free group is well-defined.

The free group on a set always exists, given by X-words modulo adjacent inversesunder concatenation with the empty word as the identity. As a corollary, everygroup is the quotient of a free group: factor the identity (set) map idG througha unique homomorphism φ : 〈G〉 → G, so that we have G ∼= 〈G〉/ kerφ since φ isepimorphic. In general, if there exist sets S of generators and R ⊆ S<ω of relationssuch that G ∼= 〈S〉/〈R〉〈S〉, we say G has presentation 〈S | R〉. We often abusenotation and write G = 〈S | R〉 when defining a group by a certain presentation.We say a group is finitely generated (f.g.) if it has a presentation with finitelymany generators. An f.g. group is finitely presented (f.p.) if it has finitely manyrelators. Given two group presentations G = 〈S | R〉 and 〈S′ | R′〉, we often write〈G,S′ | R′〉 as shorthand for 〈S ∪ S′ | R ∪R′〉. We use this contraction extensivelyin Section 4, where we prove Higman’s theorem.

Finally, we include a notational remark in the style of [3]. For presentationsG = 〈S | R〉 and G′ = 〈S′ | R′〉, we write G ≤ G′ if S ⊆ S′, R ⊆ R′, and for everyS-word w, w = 1 in G iff w = 1 in G′. Notice the condition G ≤ G′ holds iff thehomomorphism G → G′ taking an S-word in G to “itself” in G′ is an embedding.In this case we may identify G with a subgroup of G′—namely, with its image underthe described homomorphism.

2.1. Amalgamated Free Products. Recall the free product A1 ∗ A2 of disjointgroups A1, A2 satisfies the universal property that for every group G and homo-morphisms fi : Ai → G there exists a unique homomorphism φ : A1 ∗ A2 → Gthrough which the fi’s factor. Note the requirement that A1 and A2 be disjoint isnot a restriction, as we can always take disjoint isomorphic copies. The free productalways exists: if A1 = 〈S | R〉 and A2 = 〈S′ | R′〉, then A1 ∗ A2 has presentation〈S, S′ | R,R′〉. We regard amalgamated free products as a sort of generalization offree products in which we identify isomorphic subgroups.

Definition 2.1. Consider groups A1, A2 with isomorphic subgroups B1, B2 via θand inclusions i and j. The amalgamated free product A1 ∗θ A2 of A1 andA2 along θ is the group U along with homomorphisms λi : Ai → U satisfying thefollowing universal property: for every group G and homomorphisms fi : Ai → Gsuch that f1i = f2jθ, there exists a unique homomorphism F : U → G making the

4 ALEX BURKA

following diagram commute:

B1i //

jθ

A1

λ1 f1

A2λ2 //

f200

U

F G

Specifying the amalgamated free product by a universal property guaranteesuniqueness up to isomorphism, if it exists. This is in fact the case:

Proposition 2.2. Let A1, A2 be groups with isomorphic subgroups B1, B2 via θ,and let X be the set bθ(b)−1 : b ∈ B1. Then:

(a) The amalgamated free product A1 ∗θ A2 exists, given by the quotient of the freeproduct A1 ∗A2 by the conjugate closure of X along with maps λi = νµi, whereµi is the inclusion Ai → A1 ∗A2 and ν is the canonical surjection from the freeproduct A1 ∗A2 to its quotient by the conjugate closure of X.

(b) If Ai has presentation 〈Si|Ri〉 for i = 1, 2, then A1 ∗θ A2 has presentation

〈S1 ∪ S2 | R1 ∪R2 ∪X〉

Proving Proposition 2.2a involves little more than constructing an appropriatehomomorphism F and checking that the diagram commutes, as described above.Proving Proposition 2.2b follows almost immediately from our definition of grouppresentation defined in the prelude to this section. We thus leave the proof as anexercise to the willing reader; the unwilling reader may take this description as thedefinition of the free product with amalgamated subgroup.

The presentation of an amalgamated free product as the quotient of A1 ∗A2 bythe conjugate closure of X in A1 ∗ A2 concretizes the intuitive notion of “gluing”A1 and A2 together along isomorphic subgroups B1 and B2. This intuition lendsitself well to computing fundamental groups, which for certain well-behaved spaceswritten as the union of well-behaved subspaces X,Y with well-behaved intersectionis given by the amalgamated free product of π1(X) and π1(Y ) along π1(X ∩ Y ).

We now state and prove the normal form property of amalgamated free products.Recall a left transversal of a subgroup K ≤ G is a subset T ⊆ G consisting of achosen representative from the left cosets of K in G. Observe that if T is a lefttransversal, then G is the disjoint union of left cosets tK for all t ∈ T , so that everyg ∈ G has a factorization g = tk for a unique t ∈ T , k ∈ K. For our groups Ai, Bi,choose a left transversal Ti such that the representative of the trivial coset Bi is theidentity. Denote the chosen representative of akBi by ak, so that for all ak ∈ G,there exists a unique bk ∈ Bi satisfying ak = akbk.

Theorem 2.3. (Normal Forms for AFPs) In the amalgamated free product A1 ∗θA2, every element uniquely factors into a product a1 · · · anb satisfying:

(i) Each ai is a member of T1 or T2.(ii) aj ∈ A1 iff aj+1 ∈ A2.

(iii) b is an element of B1.

Such an element is called a normal form in the amalgamated free product. More-over, the maps λi : Ai → U are monomorphisms. Finally, we have U given by

HIGMAN’S EMBEDDING THEOREM AND DECISION PROBLEMS 5

the conjugate closure of λ1(A1) ∪ λ2(A2), and the intersection λ1(A1) ∩ λ2(A2) isisomorphic to B1.

Proof of Existence. Recall the amalgamated free product A1 ∗θ A2 is isomorphicto the quotient of the free product A1 ∗ A2 by the conjugate closure N of the setbθ(b)−1 : b ∈ B1 as stated in Proposition 2.2a. Each coset of N in A1 ∗ A2 hasa representative p1q1 · · · pnqn in the free product, where pi ∈ A1 and qi ∈ A2. Toshow the existence of normal forms for amalgamated free products, we proceed byinduction on the length n of these coset representatives.

For n = 1, our description above furnishes a unique b1 ∈ B1 for which p1 = a1b1.But in the amalgamated free product A1 ∗θ A2, we have identified members of B1

with their images under θ. This identification allows us to write in U

p1q1 = a1b1q1 = a1(θ(b1)q1).

But θ(b1)q1 ∈ A2 gives us another expression θ(b1)q1 = a2b2 for some uniqueb2 ∈ B2. Since b2 ∈ B2, we have b′2 := θ−1(b2) ∈ B1. Hence we see p1q1 = a1a2b

′2

in U , completing the base case.Now, assume for n− 1 we have

p1q1 · · · pn−1qn−1 = a1 · · · akbwhere the right-hand term is a normal form. Consider a word of length n of theform p1q1 · · · pnqn. Then by the inductive hypothesis, there holds

p1q1 · · · pnqn = (a1 · · · akb)pnqn = (a1 · · · ak)((bpn)qn).

But b is in B1, so bpn is in A1. Hence (bpn)qn = (ak+1b′)yn for some b′ ∈ B1.

Using the identification θ(b) = b in the amalgamated free product, this latter termequates to ak+1(θ(b′)yn). Then θ(b′)yn = ak+2b

′′ for some b′′ ∈ B2, so again weend up with an expression ak+2b

′′ = ak+2θ−1(b′′) =: ak+2bk+2. Therefore we have

p1q1 · · · pnqn = a1 · · · ak+2bk+2,

as desired. Existence of normal forms in amalgamated free products follows.

For each coset (p1q1 · · · pnqn)N of N in A1 ∗ A2, denote the normal form con-structed above by α(p1q1 · · · pnqn), so that there holds

α(p1q1 · · · pnqn)N = (p1q1 · · · pnqn)N.

Proof of Uniqueness. Denote the set of all normal forms Ω. For each a ∈ Ai,let φa : Ω → Ω be defined by φ(a1 · · · akb) = α(aa1 · · · anb). Notice φ1 = idΩ

and φa1 φa2 = φa1a2 , thus φa−1 = φ−1a . These conditions show that the maps

φa : a ∈ Ai are a subset of the group of permutations SΩ on Ω for i = 1, 2.Moreover, the map Φi : Ai → SΩ : a 7→ φa is a homomorphism. The universalproperty of free products furnishes the existence of a unique homomorphism

χ : A1 ∗A2 → SΩ : a1 · · · anb 7→ φa1 · · · φan φbmaking the diagram commute:

A1 ∗A2

χ

A1

i77

Φ1

// SΩ A2

jgg

Φ2

oo

6 ALEX BURKA

Observe for all b in B1 and all normal forms w

χ(bθ(b)−1)[w] = φbθ(b)−1(w) = α(bθ(b)−1w) = w

since b equals θ(b) in the amalgamated free product. Therefore we have bθ(b)−1 ∈kerχ for all b ∈ B1. Now, consider

Ψ : A1 ∗θ A2 → SΩ : a1 · · · anbN 7→ χ(a1 · · · anb),which is well-defined since bθ(b)−1 is in kerχ for all b ∈ B1. Two normal forms areequal iff they have the same spelling, and so Ψ(w) 6= Ψ(v) in SΩ when w 6= v fornormal forms w, v. Uniqueness thus follows.

Let i = 1, 2. Observe Ψλi(ai) = Ψ(aiN) = Ψ(aibN) = φa1 φb; one can checkai 6= 1 implies φa1 φb 6= id, so that Ψλi is monomorphic. Thus λi is monomorphic.The final results follow easily from the uniqueness of normal forms.

The following corollary essentially rephrases the Normal Form Theorem in amore tractable way. Consult Lemma 1 in [3] for an explicit proof.

Corollary 2.4. Let U = A1 ∗A1∩A2A2. If x1, ..., xn is in U \ (A1 ∩A2) and there

holds xi ∈ A1 iff xi+1 ∈ A2, then the product x1 · · ·xn is not in A1 ∩ A2. Asa consequence of the amalgamation, this implies x1 · · ·xn 6= 1 in U (for clearlyx1 · · ·xn = 1 in U implies x1 · · ·xn ∈ A1 ∩A2).

2.2. HNN Extensions. In an amalgamated free product, we essentially take afree product but identify isomorphic subgroups. HNN extensions serve a differentpurpose: in an HNN extension, we treat isomorphic subgroups as conjugate withrespect to a chosen isomorphism. In turn, the HNN extension induces this isomor-phism between these subgroups as an inner automorphism. The first part of thefollowing theorem makes sense of this notion and originally motivated HNN exten-sions, so we include it before defining them. The second part will become crucialin proving Britton’s Lemma by allowing us to transition from a special case to thegeneral theorem.

Theorem 2.5. Let G be a group.

(a) If G has isomorphic subgroups H1, H2 via some φ, there exists a group K ≥ Gcontaining some element p such that φ(a) = p−1ap for all a ∈ H1, so that φ isinduced by an inner K-automorphism.

(b) Suppose G has presentation 〈S | R〉, and let P be a set disjoint from S indexedby I. Further suppose H is a group with presentation

〈S, P | R, p−1i,jAjpi,j = Bj : ij ∈ I, j ∈ J〉

for some S-words Aj , Bj indexed by J . Define J(i) to be the set consisting of j

where p−1i,jAjpi,j = Bj as in the above presentation. If there exist isomorphisms

φi : 〈Aj : j ∈ J(i)〉 → 〈Bj : j ∈ J(i)〉 : Aj 7→ Bj

for all i, then we have G ≤ H.

HIGMAN’S EMBEDDING THEOREM AND DECISION PROBLEMS 7

Proof. To prove (a), consider the subgroups K,L generated by G ∪ q−1Aq andG ∪ r−1Br in G ∗ 〈q〉 and G ∗ 〈r〉, respectively. Then K,L are clearly free on thedisplayed generating sets. Define φ : q−1aq 7→ r−1φ(a)r and consider the diagram:

K

Θ

G

k77

iid //

id

L Aq

lgg

jφoo

φ

G

i

GG

Br

j

XX

The universal property of free products gives rise to the map Θ making the diagramcommute. In particular, Θ k = i id. Notice the generating sets of K,L have thesame cardinality. Thus Θ is an isomorphism between K,L. Moreover,

Θ(q−1aq) = jφ(q−1aq) = r−1φ(a)r.

Set H = (G ∗ 〈q〉) ∗Θ (G ∗ 〈r〉), so that H contains some isomorphic copy X of K asa subgroup by Theorem 2.3. Now, we have for all a in A, q−1aq = r−1φ(a)r in Has a consequence of the amalgamation along Θ. Setting p = qr−1 thus completesthe proof.

We can apply part (a) sufficiently many times to G so that there exists a groupH ′ containing G with generators S, pi : i ∈ I such that p−1

i,jAjpi,j = Bj holds for all

i ∈ I and j ∈ J(i). Thus there exists a surjective homomorphism H → H ′ underwhich the members of H are mapped to “themselves” in H subject to new relationsfurnished by the kernel of the map. In particular, any S-word W is the identity inH iff W is the identity in H ′ (under the homomorphism) iff W is the identity in G(as a subgroup of H ′).

For the discussion to follow, suppose G,H satisfy the hypotheses of Theorem 2.5b.

Definition 2.6. We say H is an HNN extension of G with basis S and stableletters pi if for all i ∈ I, there exists an isomorphism

φi : 〈Aj : j ∈ J(i)〉 → 〈Bj : j ∈ J(i)〉 : Aj 7→ Bj .

We now prove Britton’s Lemma. Our proof mostly follows Britton’s originalexposition in [3], but expands on the less obvious steps and moreover breaks upthe exposition into three stages. In Lemma 2.8, we consider one stable letter andassume the isomorphism φ is the identity. In Lemma 2.9, we treat isomorphismsother than the identity map but continue to restrict ourselves to one stable letter.Then, in Theorem 2.10, we prove the general case. For Lemmas 2.8 and 2.9, let Abe the subgroup in G generated by Aj : j ∈ J and B the subgroup in G generatedby Bj : j ∈ J. Similarly, , let A(i) and B(i) be the subgroups of G generated byAj and Bj for j ∈ J(i), respectively.

Definition 2.7. Suppose H is an HNN extension of G and i ∈ I. Let e = ±1. Wesay peiCp

−ei is a pinch in H if e = −1 implies C, as an element of the group G,

is an element of A(i) and e = +1 implies C is an element of B(i). A word on H’sgenerators is pi-reduced if it contains no pinches involving pi.

8 ALEX BURKA

Lemma 2.8. Let H = 〈S, p | R, p−1Ajp = Aj : j ∈ J〉 for some S-words Aj. If Wis an S, p-word involving p such that W reduces to the identity, then W contains apinch p−1Cp or pCp−1.

Proof. To prove the lemma, we first show H is isomorphic to an amalgamated freeproduct. Then we proceed by induction on the number of p-separated S-wordsneeded to describe W , using the fact that H is an amalgamated free product andapplying Corollary 2.4. To begin, we take some isomorphic copy Q1 of A via anisomorphism φ, so that Q1 has presentation

〈φ(Aj) : j ∈ J | qk = 1 : k ∈ K〉for some φ(Aj)-words qk. Take the direct product Q2 = Q1 × 〈t〉 for a letter tdisjoint from Q1. According to Proposition 2.2b, the amalgamated free productQ3 = G ∗φ Q2 has presentation

〈S, φ(Aj) : j ∈ J, p | R, qk = 1 : k ∈ K, p−1φ(Aj)p = φ(Aj), φ(Aj) = Aj : j ∈ J〉= 〈S, p | R, p−1Ajt = Aj : j ∈ J, rk := φ−1(qk) = 1 : k ∈ K〉.

The latter presentation follows from the relations φ(Aj) = Aj . Now, by construc-tion, we have qk = 1 in Q3, but in particular qk = 1 always holds in Q1. It follows,rk = φ−1(qk) = 1 in Q0. But Q0 is contained in G, so that rk = 1 holds in G. And,considering the canonical homomorphism G → Q3 formed by the composition ofthe inclusion into the (un-amalgamated) free product and the quotient map, we seerk = 1 in G implies rk = 1 in Q3. We conclude rk = 1 in G iff rk = 1 in Q3, whichallows us to omit these relations from the second presentation of Q3. With this, itbecomes apparent that we have Q3

∼= H.Since we have Q3

∼= H and G is a subgroup of Q3, it follows that G is isomorphicto a subgroup of H. In particular, an S-word w is the identity in G iff it is theidentity in H. Already the generating and relating sets in the presentation of Hcontain those of G. Altogether, we have G ≤ H.

Let W be an unreduced S, p-word involving t. We may assume W has spelling

W0pe1W1 · · · penWn

for nonzero integers ej and S-words Wj satisfying the implication [Wj is emptyimplies j = 0 or j = n]. Indeed, there is nothing to prove if W contains as asubword pp−1 or p−1p. We proceed by induction on n. Note n = 1 is impossible:the hypothesis W = 1 in H implies W = 1 in G and W = 1 in Q2. Thenpe1 = W−1

0 W−11 in G ∩Q2 follows. By Theorem 2.3, G ∩Q2

∼= A holds due to theamalgamation Q3 of G and Q2 along A. But recall A was generated by S-wordsAj , hence prohibiting the membership of pe1 .

Now assume the result for n−1, and consider a wordW spelledW0pe1W1 · · · penWn

as described above. Since there holds W = 1 in H, the contrapositive of Corollary2.4 admits the existence of j with 1 ≤ j ≤ n−1 such that we have Wj ∈ A ∼= G∩Q2.We only need to consider the case when the signs of ej and ej+1 agree, for the proof iscomplete when they do not. In this case, W contains as a subword pejWjp

ej+1Wj+1.Because Wj is a member of A, it commutes with p: the relation p−1Ajp = Aj in Hmeans pX = Xp for a product X of some Aj . Thus in H, W contains a subwordpej+ej+1WjWj+1. Having equated W to a term in H with n − 1 p and S-wordfactors, we have returned to the inductive hypothesis.

HIGMAN’S EMBEDDING THEOREM AND DECISION PROBLEMS 9

Lemma 2.9. Let H = 〈S, p | R, p−1Ajp = Bj : j ∈ J〉 for some S-words Aj , Bj.Suppose H is an HNN extension of G with basis S and stable letter p, so that thereexists an isomorphism φ : A→ B : Aj → Bj. If W is an S, p-word involving p andW reduces to the identity in H, then W contains a pinch.

Proof. Consider the following presentations:

H1 =〈S, q | Rq−1Ajq = Bj : j ∈ J〉,H2 =〈S, q, r | R, q−1Ajq = Bj , r

−1Bjr = Bj : j ∈ J〉,H3 =〈S, q, r, p | R, q−1Ajq = Bj , p = qr, p−1Ajp(= r−1q−1Ajqr) = Bj : j ∈ J〉.

Notice the isomorphisms H1∼= H and H2

∼= H3. Moreover we have G ≤ H byTheorem 2.5b. Now take an S, p-word W as in the hypothesis. As in the proof ofLemma 2.8, we only need to consider when W is spelled

W0pe1W1 · · · penWn.

Define V to be the word spelled

W0(qr)e1W1 · · · (qr)enWn.

Since the generating and relating sets of H3 contain those of H, and since W = 1holds in H, we get V = 1 in H3 due to the relation p = qr. V , H1, and H2 satisfythe hypotheses of Lemma 2.8, so that V must contain a pinch r−1Cr or rCr−1

where C ∈ B in H1. Analyzing the spelling of V , we have in the first case, C hasspelling q−1Wiq for some 1 ≤ i ≤ n − 1, so q−1Wiq is in B hence Wi is in qBq−1

hence Wi is in A. In the second case, C already has spelling Wi.

Theorem 2.10. (Britton’s Lemma) Suppose H is an HNN extension of G. If aword W involves a stable letter and W = 1 in H, then W contains a pinch.

Proof. Suppose H has presentation

〈G, pi : i ∈ I | p−1i,jAjpi,j = Bj : j ∈ J(i)〉.

Order the elements of I (it is countable) and consider the following:

H0 = 〈G〉, Hk = 〈G, pk | p−1k,jAjpk,j = Bj : j ∈ J(k)〉.

Since H is an HNN extension of G, there are isomorphisms φi : A(i)→ B(i) for alli. We thus have Hk+1 is an HNN extension of Hk for all k with basis

S ∪ pi : 1 ≤ i ≤ k − 1and stable letter pk, as the isomorphism condition is already fulfilled. Thus, weobtain a chain Hk ≤ Hk+1 and Hk ≤ H for all k by Theorem 2.5b. Define N =maxn ∈ N : W involves pn; we proceed by induction on N . Lemma 2.9 coversthe case N = 1. Assume the result for N − 1. Given the hypothesis W = 1 in H,the chain H0 ≤ H1 ≤ · · · , and the maximality of N , it follows, W = 1 in HN . ByLemma 2.9, W contains a subword p−1

N CpN or pNCp−1N where in the first case, C is

a member of the subgroup 〈Bj : j ∈ J(N)〉HN−1 and in the second, C is a memberof the subgroup 〈Aj : j ∈ J(N)〉HN−1 . The proof is complete if C is an S-word,for the aforementioned subgroups in HN−1 are subgroups in G as well. If not, Cinvolves some p1, ..., pN−1 equating in EN−1 to a word D in B(i) (in G). Thatis, there holds CD−1 = 1 in EN−1. By the inductive hypothesis, CD−1 containsa subword of the desired form. Since D does not contain any pi, we know thissubword occurs in C. Then, it also occurs in W .

10 ALEX BURKA

3. Turing Machines and Associated Semigroups

Having established the basic combinatorial group-theoretic notions of amalga-mated free products, HNN extensions, and their normal forms, we shift our at-tention to the necessary topics in computability theory. Albeit self-contained, ourtreatment of computability does not motivate the definitions to follow. It insteadaims to make apparent why the associated semigroup behaves like a Turing machine(Lemma 3.5). Turing’s original paper [10] alongside a more modern treatment likeCooper’s [4] are good sources for understanding the intuition behind the elementarymaterial of computability theory.

Definitions 3.1 and 3.2 and Proposition 3.3 comprise this elementary material.We include Definition 3.1 in full because the way we formulated it facilitates ourdefinition of the associated semigroup. The proof of Proposition 3.3 is standardand elementary, so we omit a full treatment from our discussion aside from a fewbrief remarks. Refer to [10] and [4] for an explicit proof.

Definition 3.1. Let S be a finite set of symbols s0, ..., sm and Q a finite set ofstates q0, ..., qn; s0 is the blank symbol, and q1 is the initial state. We denote theset of quadruples Q × S × S ∪ L,R × Q by Q, where L,R is disjoint from Qand S. For convenience, we identify quadruples in Q with strings. We say a setT ⊆ Q of quadruples is consistent if for all q1

i s1js

1kq

1l and q2

i s2js

2kq

2l in Q, q1

i s1j = q2

i s2j

implies s1kq

1l = s2

kq2l . A deterministic Turing machine T is a finite, consistent

set of quadruples in Q. As far as potentially confusing terminology goes, we call Sthe alphabet of T , Q the states of T , and S ∪Q ∪ L,R the symbols of T .

A description of a Turing machine T is a finite string si1 · · · sikqjsik+1· · · sil of

symbols and exactly one state not occurring at the right end. For descriptions X,Yof T , we have the basic move X → Y if one of the conditions below holds whereinP,Q are S-strings:

qisjskql ∈ T ∧ (X = PqisjQ ,Y = PqlskQ);

qisjRql ∈ T ∧ (X = PqisjstQ,Y = PsjqlstQ) ∨ (X = Pqisj , Y = Psjqls0);

qisjLql ∈ T ∧ (X = PsjqistQ,Y = PqlsjstQ) ∨ (X = qisjQ,Y = qls0sjQ).

The stipulation that T be a consistent subset of Q prohibits non-deterministicmoves, where X → Y and X → Z but Y 6= Z.

Let S∗ denote the set of finite strings of S-letters. Given a string σ in S∗ whosefirst character is not blank, the description q1σ is called an input to T . A terminalcomputation T (σ) with input q1σ is a finite collection of descriptions X0, ..., Xp

ordered such that q1σ = X0 → · · · → Xp and for no i < p does there hold Xp → Xi.Write T (σ) ↓ if there exists a terminal computation with input q1σ. The haltingset h(T ) of T is the set σ ∈ S∗ : T (σ) ↓; we say T enumerates the set h(T ). Wesay q0 is a stopping state of T if for every σ ∈ h(T ), the final description of thecomputation beginning with q1σ involves the state q0. Note we may assume anyTuring machine has stopping state q0 without altering its halting set.

A subset E ⊆ S∗ is computably enumerable (c.e.) if there exists a Turing ma-chine T halting on inputs from E—that is, if we have E = h(T ). E is computableif E and Ec are c.e.

HIGMAN’S EMBEDDING THEOREM AND DECISION PROBLEMS 11

We may now ask the question, does there exist a c.e. but not computable set?The following construction shows this is the case: Let S = s0, s1 and Q be anyfinite set of states. Here, s0 is interpreted as 0 and s1 is interpreted as 1; stringsof n + 1 s1’s represent natural numbers n via a tally encoding. Observe that thenumber of Turing machines on any finite alphabet and finite state set is countable.The fact that there are countably many Turing machines admits the existence of alisting T0, T1, ..., Tn, ... of all Turing machines on S and Q.

Definition 3.2. Define the Halting Problem to be the set of natural numbers non which the n-th Turing machine halts:

0′ := n ∈ N : Tn(sn+11 ) ↓.

Proposition 3.3. 0′ is c.e. but not computable.

Proof. If [0′]C were c.e., then Tk enumerates [0′]C for some k. The contradictionk ∈ 0′ ↔ k ∈ [0′]C follows. But 0′ is c.e., for the universal Turing machineU enumerates 0′. Its movement reminisces the Cantor snake along the infinite 2Darray with n-th column the (possibly nonterminal) computation Tn(sn+1

1 ).

We end our brief discussion of Turing computability in its own right and shift ourfocus to the associated semigroup. Definition 3.1 easily translates to a semigroupcharacterization: the quadruples in the machine correspond to the relations imposedon the semigroup generated by the symbols and state set. We include additionalsymbols h to represent the ends of the tape and q to represent a sort of terminalstate. These additional symbols and relations as described in the definition allowus to prove Lemma 3.5, the key result of this section. This lemma establishes alink between the set enumerated by the machine and an algebraic condition in thesemigroup, showing our semigroup behaves much like the machine after all.

Before defining the associated semigroup and proving Lemma 3.5, we include anotational remark. Fix a generating set X and consider the semigroup S generatedby X subject to the relations Aj = Bj for some X-words Aj , Bj indexed by a setJ . For X-words W,V and P,Q, we write W → V if W and V are spelled PAjQand PBjQ for some j, or vice-versa. Observe two X-words W,V are equal in S iffthere are finitely many U1, ..., Un such that there holds

W ≡ U1 → · · · → Un ≡ V

Definition 3.4. Let T be a Turing machine with stopping state q0 and symbolsL,R, q0, ..., qN , s0, ..., sM. Define the associated semigroup Γ(T ) by the fol-lowing presentation:

Generators G(T ) Relations R(T )

q, h qisj = qlsk if qisjskql ∈ T

q0, ..., qN qisjsb = sjqlsb if qisjRql ∈ T

s0, ..., sM qisjh = sjqls0h if qisjRql ∈ T

sbqisj = qlsbsj if qisjLql ∈ T

hqisj = hqls0sj if qisjLql ∈ T

q0sb = q0, sbq0h = q0h, hq0h = q

12 ALEX BURKA

Lemma 3.5. Let T be a Turing machine with stopping state q0, let W be the setof positive words on the alphabet S of T , and let E be the set enumerated by T . Forany w ∈W , we have w ∈ E iff hq1wh = q in the associated semigroup Γ(T ).

Proof. Suppose w ∈ W and hq1wh = q in Γ(T ). By the note above, there areelementary moves U0, ..., Un in Γ(T ) so that there holds

hq1wh ≡ U0 → · · · → Un ≡ hq0h→ q.

Notice that if U, V are words in Γ(T ) with U → V , and neither U nor V is spelledq, then we have U ≡ hXh iff V ≡ hY h for some descriptions X,Y of T ; for theonly operation in Γ(T ) killing h is hq0h = q, thus the observation follows from thehypothesis U, V 6≡ q. In particular, each Uk is of the form hXkh for a descriptionXk. Let m be the first k such that Uk involves q0; we proceed by induction onm. Clearly m ≥ 1, since X0 involves q1 and no other qi. Observe that if we haveU ≡ hXh for a description X, V 6≡ q, and U → V in Γ(T ) describes one of the firstfive relations, then V ≡ hY h for a description Y such that X → Y or Y → X in T .Thus for all i, we have Xi → Xi+1 or Xi+1 → Xi. But Xn involves stopping stateq0, hence Xn−1 → Xn. For m = 1, we recover q1w ≡ X0 → X1 as a computationin T , verifying T (w) ↓. Assume the result for all m < M for M > 1. If we have

X0 → X1 → · · · → XM−1 → XM ,

we are done. Otherwise, there exists k such that Xk → Xk−1 and Xk → Xk+1.But since T is deterministic, this implies Xk−1 ≡ Xk+1, hence reducing to the

case M − 1. Relabelling the descriptions X1, ..., Xk, ...Xn to X0, ..., XM−1, thecomputation T (w) must halt by the inductive hypothesis.

The forward implication is easier. If w is in E, there is a computation

T (w) ≡ X0 → · · · → Xn.

Notice how the first five relations in Γ(T ) arise from the basic moves of the machineoutlined in Definition 3.1. Repeated application gives us

hq1wh(= hX0h) = hX1h = · · · = hXnh.

Since T has stopping state q0, Xn must involve q0—that is, there holds Xn ≡ Fq0Gfor S-words F,G. But for all sk ∈ S, we have q0sk = q0 and so there holds in Γ(T )

hXnh = hFq0Gh = hFq0h.

Similarly, the relations skq0h give us in Γ(T )

hFq0h = hq0h = q.

That is, hq1wh = q holds in Γ(T ), as desired.

4. Higman’s Embedding Theorem

In this section we prove Higman’s Embedding Theorem, the main result of thispaper. However, our proof does not follow Higman’s original 1961 paper but re-lies on [1]. Higman developed a theory of “benign subgroups” to reach the re-sult, whereas Aanderaa uses the theory we have already developed in the last twosections—namely, the normal form properties of amalgamated free products andHNN extensions, and the correspondence in Lemma 3.5.

How can we characterize the f.g. subgroups of an f.p. group? As Higman’stheorem shows, they are exactly the recursively presented subgroups.

HIGMAN’S EMBEDDING THEOREM AND DECISION PROBLEMS 13

Definition 4.1. An f.g. group is recursively presented if there exists a presenta-tion 〈s1, ..., sM | w = 1 : w ∈ R〉 where R is a recursively enumerable set of positivewords on the generators.

Requiring that R consist of positive words on the generators is not a restriction.For any group presentation 〈S | R〉 and for every s ∈ S such that s−1 occurs ina word in R, adjoin a new generator t, replace the occurrences of s−1 with t inthe relevant relations of R, and adjoin the relation ts. Call the new presentation〈S′ | R′〉; the presentations 〈S | R〉 and 〈S′ | R′〉 evidently define the same group.This condition will simplify how we translate between semigroups and groups.

We now state Higman’s Theorem, but delay its proof until the end of the sec-tion. The proof of sufficiency is rather involved, requiring a technical lemma thatestablishes a correspondence between the associated semigroup of a Turing machineand a group Q∗, whose definition will come. The proof of this lemma will use anoft-used fact which we include beforehand.

Theorem 4.2. (Higman, 1961) An f.g. group G is recursively presented iff thereexists an f.p. group H into which G embeds.

Before proving Higman’s theorem, we must introduce a series of group presen-tations. Defining these presentations will rely on the following notational remarks.We fix this notation for the remainder of the section. In particular, when referringto Γ(T ) in Lemma 4.8, we refer to Γ(T ) as described below.

Notation 4.3. We begin with a recursive presentation

〈u1, ..., um | w = 1 : w ∈ R〉of a group G. By definition, there exists a Turing machine T with alphabets0, ..., sM containing the generators of G and states q0, ..., qN with q0 a stoppingstate q0, such that T enumerates R. For convenience, we abbreviate the relatingset of Γ(T ) by expressions Fiqi1Gi = Hiqi2Ki for some positive words on the al-phabet s1, ..., sM , h and qi1 , qi2 ∈ q, q0, ..., qN indexed by a set I. For a wordX ≡ xj1 ...xjn , we write X to denote the word spelled x−1

j1...x−1

jn.

Presentations 4.4.

G =〈u1, ..., um | w = 1 : w ∈ R〉,Q0 =〈x〉,Q1 =〈Q0, h, sk : 1 ≤ k ≤M | xsk = skx

2;xh = hx2〉 ,

Q′1 =Q1 ∗ 〈q, q1, ..., qN 〉,Q2 =〈Q1, ri : i ∈ I, q, ql : 1 ≤ l ≤ N | risk = skxrix; rih = hxrix; r−1

i F iqi1Giri = Hiqi2Ki〉,Q3 =〈Q2, t | tri = rit, tx = xt〉,Q′3 =〈Q2, t0 | t−1

0 (q−11 hrih

−1q1)t0 = q−11 hrih

−1q1; t−10 (q−1

1 hxh−1q1)t0 = q−11 hxh−1q1〉,

Q∗ =〈Q3, k | kri = rik; kx = xk; k(q−1tq) = (q−1tq)k〉,Q′∗ =〈Q′3, k0 | k−1

0 (hrih−1)k0 = hrih

−1, k−10 (hxh−1)k0 = hxh−1;

k−10 (hq−1h−1q1t0q

−11 hgh−1)k0 = hq−1h−1q1t0q

−11 hqh−1〉,

Q4 =Q′∗ ∗G,

14 ALEX BURKA

Q5 =〈Q4, bs : 1 ≤ s ≤ m | b−1s ujbi = uj ; b

−1s ajbs = aj ; b

−1s k0bs = k0u

−1s : 1 ≤ s, j ≤ m〉,

Q6 =〈Q5, d | d−1k0d = k0, d−1asbsd = as : 1 ≤ s ≤ m〉,

Q7 =〈Q6, e | e−1t0e = t0d; e−1k0e = k0, e−1ase = as : 1 ≤ s ≤ m〉.

Lemma 4.5. Q1 is an HNN extension of Q0. Q2 is an HNN extension of Q′1. Q3

and Q′3 are isomorphic and are HNN extensions of Q2. Q∗ and Q′∗ are isomorphicand are HNN extensions of Q3 and Q′3, respectively. Q5 is an HNN extension ofQ4, Q6 is an HNN extension of Q5, and Q7 is an HNN extension of Q6.

Remark 4.6. We omit proofs of the claims in Lemma 4.5 from our discussion, asthey are uninteresting in their own right. Full proofs can be found in [1] and [8].Note that, albeit arid, these proofs are not all trivial. In particular, showing Q7

is an HNN extension of Q6 is especially difficult. Though we do not verify theseHNN extensions, we will use their normal form properties extensively to arrive atthe desired result.

Lemma 4.7. Suppose H is an HNN extension of G with stable letters ri : i ∈ I,and suppose W and V are ri-reduced words for all i of the form

W0re1i W1 · · · remi Wm, V0r

f1i V1 · · · rfni Vn

where Wj , Vj do not involve any stable letters for all 1 ≤ j ≤ m,n respectively.If W = V in H, then we have m = n, ej = fj for all j. Furthermore, the word

remi WmV−1n r−fni is a pinch.

Proof. We proceed by induction on maxn,m. By hypothesis we have WV −1 = 1in H, hence Britton’s lemma guarantees a pinch in WV −1. With W,V ri-reduced

for all i, the pinch must occur at the interface remi WmV−1n r−fni , as claimed. Hence

we must have em = fn since em and −fn differ in sign. The lemma clearly holdsfor maxm,n = 1. Assume the result for maxm,n = N − 1, and consider thecase maxm,n = N . If em = fn = −1, then by the definition of a pinch, Britton’slemma asserts WmV

−1n = Ag11 · · ·A

gmk where Aj ∈ A(ri) and gj = ±1 for 1 ≤ j ≤ k.

Then our original pinch r−1i WmV

−1n ri equates in H to

r−1i Ag11 · · ·A

gmk ri = r−1

i Ag11 rir−1i · · · r

−1i Agmk ri,

which in turn equates in H to Bg11 · · ·Bgkk for some Bj ∈ B(ri). Having eliminated

one ri from the pinch r−1i WmV

−1n ri, we have reduced this case to the inductive

hypothesis. A similar argument holds for the case em = fn = 1.

Lemma 4.8. Let w be a positive word on the alphabet of T , and define

τ = h−1q1wh; σ = hq1wh; Λ = k−1(τ−1tτ)k(τ−1t−1τ).

Then we have σ = q in Γ(T ) iff there holds Λ = 1 in Q∗.2

Proof of Necessity in Lemma 4.8. In addition to the hypothesis Λ = 1 in Q∗, wehave Q∗ is an HNN extension of Q3 and Λ contains the stable letter k of Q∗. ThusBritton’s lemma guarantees a pinch k−1Ck for a word C on some ri, x, and q−1tq.Inspecting the form of Λ shows us C = τ−1tτ in Q3. That is, there holds

τ−1tτ = R−1n (q−1tq)en · · ·R−1

1 (q−1tq)e1R−10

2That is, iff τ−1tτ commutes with k in Q∗.

HIGMAN’S EMBEDDING THEOREM AND DECISION PROBLEMS 15

for some x, ri-words Rj and ej = ±1. Let n be such that this equation involves theminimal number of terms. Now, let M be the word with spelling

τ−1tτR0(q−1te1)R1(q−1te2q)R2 · · · (q−1tenq)Rn

so that there holds M = 1 in Q3. Applying Britton’s lemma again, we see Wcontains a pinch tCt−1 or t−1Ct where C = R∗ in Q2 for some word R∗ on ri andx. Assume the initial t or t−1 in tCt−1 or t−1Ct, respectively, occurs in tej for somej ≥ 1. Because C does not involve t, we have

teCt−e ≡ tejqRjq−1tej+1 .

In Q3, the relations txt−1x−1 and trit−1r−1

i admit the following equations:

q−1tejqRjq−1tej+1q ≡ q−1teCt−eq

= q−1teRt−eq

= q−1Rq

= q−1(qRjq−1)q

= Rj .

We have thus contradicted the minimality of n, forbidding this case. If, in fact, theinitial t or t−1 occurs in the first possible place in W , then we have

tCt−1 ≡ tτR0q−1te1 ,

so that R−1τR0 = q holds in Q2, where R,R0 are words on ri and x. Expandingand rearranging, we obtain in Q2 the equation

R−1h−1q1 = qR−10 h−1w−1.

Notice this final equation still holds when R and R0 are freely reduced (that is,when they contain no adjacent inverses). Thus R−1h−1q1 and qR−1

0 h−1w−1 areri-reduced for all i, since w is a positive word on the alphabet of T and R,R0 arefreely reduced x, ri-words.

The remainder of this proof involves showing hq1wh = q in Γ(T ), providedR−1h−1q1 and qR−1

0 h−1w−1 are ri-reduced for all i. We proceed by induction onthe number N of ri’s in R−1, which coincides with the number of ri’s in R−1

0 byLemma 4.7. The case N = 0 is not possible.3 The base case follows a similarargument to the inductive step, so we omit a complete description here. Assumethe result for all T < N , and then assume there are N ri’s occurring in R−1 andin R0. By Lemma 4.7, we may write in Q2

R−1τR0 ≡ R1[rei xmτxnr−ei ]R2 = q,

where we have R−1 = R1rei and R0 = r−ei R2. Moreover, the bracketed term is a

pinch. By Britton’s lemma, the term xmτxn lies in

A(i) = 〈F iqi1Gi, s1x1, . . . , sMx〉Q′1 , B(i) = 〈Hiqi2Ki, s1x

−1, . . . , sMx−1〉Q

′1

3If we have N = 0, then it follows q = R−1τR0 = xmτxn in Q2 for some integers m,n. Since

Q′1 ≤ Q2 holds and no ri’s occur by hypothesis, xmτxn = q in Q′1. This equation holds in the

free product Q′1 iff q1 = q and xmh−1whxn = 1 in Q1, implying m = n = 0 and in turn implying

τ is empty, a contradiction.

16 ALEX BURKA

when e = −1 and e = 1, respectively. We will only treat the former case, the latterbeing similar. For the case e = −1, we see j = i1, qj = qi1 . Since xmτxn is in A(i),we obtain in Q′1 the equation

xmτxnW0(F iqi1Gi)f1W1 · · · (F iqi1Gi)fnWn = 1

where fj = ±1 and Wj are words on the free generators s1x, . . . , sMx. Hencewe may assume the Wj ’s are freely reduced. Define Λ′ to be the first term in theequation above, and assume it is spelled with the minimal number n of terms. Wemay further assume each Wj is freely reduced without changing the equality above.Since Q′1 is an HNN extension of Q1 and Λ′ involves a stable letter, Britton’s lemmaguarantees that Λ′ contains a pinch. If this pinch occurs at the first occurrence ofqi1 , then we must have f1 = −1 and whxnW0G

−1i = 1 in Q1. If the pinch occurs

anywhere else, say the j-th qi1 , it follows, uj = 1, a contradiction lest we violatethe minimality of n. Thus the equations f1 = −1 and

M ≡ xmh−1q1whxnW0G

−1i q−1

i1F−1

i W1 = 1

in Q′1 follow. We already showed whxnW0G−1i = 1 in Q1. Since Q′1 is a free

product, we have F−1

i W1 = 1 in Q1. Conjugating, the following holds in Q1:

xnW0G−1i wh = 1 = h−1F

−1

i u1xm.

Recall the words Wj are freely reduced, so in particular they contain no subwords

of the form sks−1k or s−1

k sk, for 1 ≤ k ≤M . Cancelling any remaining subwords ofthis form does not change any of the equalities above. After performing this reduc-tion, the first surviving letter in G−1

i wh after the reduction is positive. Because Giis a positive word on the alphabet of T , G−1

i necessarily vanishes upon free reduc-

tion. If not, G−1i wh begins with s−1

k for some k. But xnW0G−1i wh =Q1

1 involves

sk, so Britton’s lemma guarantees a pinch sekCs−ek . Since W0 is freely reduced, this

pinch cannot occur as a subword of xnW0. Thus the last letter of the pinch is thefirst surviving letter of G−1

i . If, as we assumed, this letter is not positive, then we

have e = 1 and sekCs−ek ≡ skxs

−1k , for C is an x-word. But Britton’s lemma further

deduces x ∈ 〈x2〉, a contradiction. Similarly, F−1

i vanishes in h−1F−1

i .Define V0 :≡ r−1

i W−10 ri, so that V −1

0 is a word on s1x−1, . . . , sMx

−1 due to the

relations r−1i skxri = skx

−1 for all 1 ≤ k ≤ M . Consider the automorphism ψ on

Q1 given by ψ(x) = x−1 and ψ(sk) = s−1k ; then we get ψ(W−1

0 ) = V −10 . Then we

have V −10 = X0x

−n in Q1 where X0 is a positive word on the alphabet of T withspelling G−1

i wh. Setting V1 :≡ r−1i W1ri, a similar argument gives V −1

1 = x−mX1

where X1 is a negative word on the alphabet of T with spelling h−1F−1

i . With thisinformation, we end up in Q2 with the equation

q = R1x−mX1Hiqi2KiX0x

−mR2.

But R1x−m and x−nR2 are x, ri-words having at most N−1 occurrences of various

ri. One can easily check X1Hi and KiYi are freely reduced. Hence by the inductivehypothesis, X1Hi and KiX0 are positive words with h−1 ≡ X1F1 and wh ≡ KiX0.Furthermore, the inductive hypothesis guarantees X1Hiqi2KiX0 = q in Γ(T ). Itfollows, there holds in Γ(T ) the equation

hq1wh ≡ X1FiqiGiX1 = X0Hiqi2KiY1.

That is, we arrive at the desired expression hq1wh = q in Γ(T ).

HIGMAN’S EMBEDDING THEOREM AND DECISION PROBLEMS 17

Proof of Sufficiency in Lemma 4.8. If we have σ = q in Γ(T ), there are moves

σ ≡ U0 → · · · → Un ≡ hq0h→ q

for some words U0, ..., Un in Γ(T ) of the form

Uk ≡ XFiqi1GiY ; Uk+1 ≡ XHiqi2KiY

for positive S-words X,Y . In Q∗, the relations inherited from Q2 give us

X(Hiqi2Ki)Y = X(r−1i F iqi1Giri)Y = LX(F iqi1Gi)Y R

for some words L,R on various ri and x. In Γ(T ), we have Uk = Uk+1. Thus inQ∗, we have the equation U ′k = U ′k+1 where

U ′k = X(F iqi1Gi)Y ; U ′k+1 = X(Hiqi2Ki)Y

as above. For each k, reassign the specific L and R above to Lk and Rk accordingly.Now, let L = L1 · · ·Ln−1 and R = Rn−1 · · ·R1. In Q∗, we obtain U ′1 = LU ′nR. Theequation τ = LqR in Q∗ now readily follows from U1 = σ iff U ′1 = τ and U ′n = q.Inspecting the relations in Q∗ inherited from Q3 and the relations involving k, wesee that t and k commute with x, ri for all i, hence t and k commute with L andR. In Q∗, we obtain the desired equation:

(τ−1tτ)k = R−1q−1(L−1tL)q(Rk)

= R−1((q−1tq)k)R

= (R−1k)q−1tqR

= kR−1q−1L−1tLqR

= k(τ−1tτ).

Proof of Necessity in Theorem 4.2. Identify G with its image under the embeddinggiven in the hypothesis. If H is f.p. then, a fortiori, H is f.g. and recursivelypresented. That is, 〈S | R〉 presents H for a finite set S and c.e. set R of positivewords on S. By definition, we have H ∼= 〈S〉/〈R〉〈S〉. Since R is c.e. and 〈R〉〈S〉consists of conjugates of elements of R, so too 〈R〉〈S〉 is c.e. If we have G ≤ H,then there is an isomorphism

H ∼= K/(K ∩ 〈R〉〈S〉),where K is an f.g. subgroup of 〈S〉. K must be c.e., thus so too K ∩ 〈R〉〈S〉 mustbe c.e. since the finite intersection of c.e. sets is c.e. Given

〈K ∩ 〈R〉〈S〉〉K = K ∩ 〈R〉〈S〉

we have a presentation 〈K | K ∩ 〈R〉〈S〉〉 of H. If K ∩ 〈R〉〈S〉 involves any negativeletters, produce an equivalent presentation omitting them. In any event, H has arecursive presentation.

Proof of Sufficiency in Theorem 4.2. In Lemma 4.8, we showed σ = q in Γ(T ) iff(τ−1tτ)k = k(τ−1tτ) in Q∗. In Lemma 3.5, we showed σ = q in Γ(T ) iff w ∈ R,since T enumerates R. Recall the relations

t0 = q−11 hth−1q1; k0 = hkh−1

18 ALEX BURKA

hold in Q′3 and Q′∗ respectively, and recall the isomorphisms Q′3∼= Q3 and Q′∗

∼= Q∗.If w is a positive word on the alphabet of T , then we have w ∈ R iff in Q′∗,

(∗) k0(w−1t0w) = (w−1t0w)k0.

If w is a word on a1, ..., am, let w[b] and w[u] be the words formed by replacing aiwith bi and ui, respectively. Theorem 2.5b rephrases Lemma 4.5 as follows:

Q0 ≤ Q1 −→ Q′1 ≤ Q2 ≤ Q′3 ≤ Q∗ −→ Q4 ≤ Q5 ≤ Q6 ≤ Q7.

Recall 〈S | R〉 ≤ 〈S′ | R′〉 for S ⊆ S′, R ⊆ R′ iff the homomorphism sendingan S-word to “itself” in S′ is an embedding. Now, G embeds into Q4 := Q∗ ∗ Gvia the canonical maps. Thus G embeds into Q7 by repeated applications of ourremark above. Showing Q7 has a finite presentation thereby completes the proofof sufficiency in Higman’s theorem.

To show Q7 has a finite presentation, consider the presentation Q′7 identical toQ7 in all but omitting those relations of the form w[u] = 1 for w ∈ R. It is enoughto demonstrate these omitted relations are derivable from the remaining relationsin Q′7, of which only finitely many remain. To that end, take any w ∈ R. Then:

(0) k−10 w−1t0wk0k0 = w−1t0w. .

(1) e−1(k−10 w−1t0wk0)e = e−1w−1t0we.

(2) k−10 w−1e−1t0ewk0 = w−1e−1t0ew.

(3) k−10 w−1t0dwk0 = w−1t0dw.

(4) (k0w−1t0wk0)k−1

0 w−1dwk0 = (w−1t0w)w−1dw.

(5) k−10 w−1dwk0 = w−1dw.

(6) dwd−1 = ww[b]. .

(7) w−1dw = w−1dd−1ww[b]d = w[b]d.

(8) k−10 w[b]dk0 = w[b]d.

(9) k−10 w[b]k0 = w[b], k−1

0 w[b]k0 = w[b]w[u].

(10)w[u] = 1.

(0) follows from (∗), and (1) follows immediately. Observing that e commutes withk0 and all ai in the presentation of Q7, we derive (2). Then (3) follows from therelation e−1t0e = t0d in Q7. In (4) we insert the identity wk0k

−10 w−1 on the left

and ww− on the right and then rearrange. By rearranging, (5) readily follows by(∗). The relations d−1aibid = ai and aibj = bjai give rise to (6): since w is a wordai1 · · · ain on ai in Q7, we have

dai1 · · · aind−1 = dd−1ai1bi1d · · · d−1ainbindd−1 = ai1bi1 · · · ainbin = ww[b].

Then (7) readily follows from (6), and (8) follows by combining (5) and (7). The firstpart of (9) expresses the commutativity relation on k0 and d in Q6. The relationsk0bik0 = biui and biuj = ujbi admit the second part, since if w[b] = bi1 · · · bin and

w[u] = ui1 · · ·uin , then:

k−10 bi1 · · · bink0 = k−1

0 bi1k0k−10 bi2k0 · · · k−1

0 bikk0 = bi1ui1 · · · binuin = w[b]w[u].

Combining the two parts of (9) proves w[b]w[u] = w[b], from which the desired result(10) follows. This completes the proof.

HIGMAN’S EMBEDDING THEOREM AND DECISION PROBLEMS 19

We have thus proved Higman’s theorem for groups, but we may ask the samequestion for other algebraic objects. This is precisely the content of the Boone–Lawvere Conjecture.

Question 4.9. (Boone, Lawvere) How general is Higman’s theorem? For whichalgebraic categories does there exist an analog of Higman’s theorem? Does thereexist an analog of Higman’s theorem for every single-sorted algebraic theory?

5. Applications to Decision Problems

Having presented a proof of Higman’s Theorem, we conclude this paper withsome applications to decision problems, as the title of this section suggests. Ourfirst application is the Novikove–Boone Theorem, which follows as a corollary toHigman’s Theorem. With the existence of an f.p. group with unsolvable wordproblem secured by the Novikove–Boone Theorem, we proceed to prove a muchmore powerful result—the Adian-Rabin Theorem. A special case of this theoremthen shows the homeomorphism problem for manifolds of dimension greater than3 is not decidable, bringing us full-circle with the Introduction.

5.1. The Word Problem. Novikov and Boone’s independent proofs of the Novikove–Boone Theorem in the 1950s relied on a “combinatorial tour-de-force,” as describedby Baumslag [2]. In contrast to Novikov and Boone’s dense combinatorial argu-ments, a strikingly simple proof follows almost immediately from Higman’s Embed-ding Theorem.

Definition 5.1. Let G be an f.g. group with generators x1, ..., xn, and let Wbe the set of positive words on these generators. The word problem for G issolvable if the set w ∈W : w = 1 in G is computable.

Theorem 5.2. There exists an f.p. group with unsolvable word problem.

Proof. Consider isomorphic copies 〈a, b〉, 〈c, d〉 of the free group of rank 2, and anisomorphism φ between 〈a−nban : n ∈ 0′〉 and 〈c−ndcn : n ∈ 0′〉 defined on thegenerators by φ(a−nban) = c−ndcn for all n ∈ 0′. Define G to be the amalgamatedfree product 〈a, b〉 ∗φ 〈c, d〉. Then G has presentation

〈a, b, c, d | a−nbanc−nd−1cn : n ∈ 0′〉.Inspecting the presentation shows for any word

w ≡ a−nbanc−nd−1cn

with n ∈ N, we have w = 1 in G iff n ∈ 0′. That is, G has unsolvable word problem,lest we compute 0′.

But 0′ is c.e., so G has a recursive presentation. By Higman’s EmbeddingTheorem, there exists an f.p. group H into which G embeds as a subgroup. SoH must also have unsolvable word problem.

5.2. Markov Properties and a Universal F.P. Group. The existence of anf.p. group with undecidable word problem sets the foundation for a more generalundecidability result. This result, the Adian-Rabin Theorem, claims Markov prop-erties are not decidable. As it turns out, most “reasonable” properties of a groupare Markov. Example 5.4 presents several Markov properties, but many more exist.

Once we establish the Adian-Rabin Theorem, we bridge a connection to Higman’stheorem by introducing the notion of a universal f.p. group—that is, an f.p. group

20 ALEX BURKA

into which every f.p. group embeds. Corollary 5.8 then combines these results,connecting universality (as in the sense above) and Markov properties.

Definition 5.3. Given a property P of a group, we say P is a Markov propertyif P is preserved under isomorphism, there exists an f.p. group satisfying P , andthere exists an f.p. group not embeddable into any f.p. group satisfying P .

Example 5.4. The following are Markov properties:

(1) Trivial: Only the trivial group embeds into the trivial group.(2) Finite: Z does not embed into any finite group.(3) F.G. and Free: If a group embeds into a free group, its image is free (Nielsen-

Schreier Theorem). Moreover, Markov properties are preserved under isomor-phism. Thus any non-free, f.p. group like Z2 will do.

(4) F.G. with Solvable Word Problem: Finite groups, free groups, and residuallyfinite groups are all f.p. groups with solvable word problem. Theorem 5.2furnishes an f.p. group with unsolvable word problem, and clearly no suchgroup embeds in a group with solvable word problem.

Theorem 5.5. (Adian, Rabin) Fix a countable alphabet S. For any Markov prop-erty P , the set of f.p. groups generated by S satisfying P is not computable.

Proof. By definition there exist A satisfying P and B not embeddable into anysubgroup of an f.p. group satisfying P . Let C be an f.p. group with unsolvableword problem. Then C ∗B has unsolvable word problem, and moreover has a finitepresentation 〈x1, ..., xn | R〉. Now, let w be a word on the generators of C andconsider the following group presentations:

D = (C ∗B) ∗ 〈p〉 = 〈q0 := p, q1 := px1, ...qn := pxn | R′〉,E = 〈D, r0, ..., rn | riq2

j r−1i = qj : 1 ≤ i, j ≤ jn〉,

F = 〈E, s | sris−1 = r2i : 1 ≤ i ≤ n〉,

G = 〈a, b | bab−1 = a2〉,H = 〈G, c | cbc−1 = b2〉,J = 〈E ∗H | sa−1, wq0w

−1q−10 c−1〉,

K = J ∗A.

Some remarks: R′ swaps instances of xi with p−1qi. E is an HNN extension ofD with stable letters ri. The subgroups of E generated by r0, ..., rn and r2

0, ..., r2n

are free by contraposing Theorem 2.10 and are thus isomorphic, so F is an HNNextension of E with stable letter s. H and G are HNN extensions of G and F ,respectively. If the word w is trivial in C, then following the relations in J showsus all its generators are trivial, so J is trivial. Then K = A, so K satisfies P . If wis not trivial in C, then a, c and s, wq0w

−1q−10 freely generate their subgroups in H

and F , since wq0w−1q−1

0 has infinite order in E and c is a stable letter. Thus wehave 〈a, c〉H ∼= 〈s, wq0w

−1q−10 〉 via the map

ρ : a 7→ s, c 7→ wq0wq−10 .

As suspected, there is an isomorphism J ∼= F ∗ρ H after all. Thus, F embeds intoJ , and J embeds into K. One already sees B embeds into F by construction. SinceB then embeds into J , J cannot satisfy P by definition. Thus K does not satisfy P .

HIGMAN’S EMBEDDING THEOREM AND DECISION PROBLEMS 21

We have shown w = 1 in C iff K satisfies P . But C was taken to have undecidableword problem, so we cannot decide whether K satisfies P .

Corollary 5.6. The isomorphism problem is not computable.

Proof. Assume for contradiction we could decide G ∼= H for any f.p. groups G,K.Then we could decide G ∼= 1, a contradiction since triviality is Markov.

Theorem 5.7. There exists an f.p. group containing an isomorphic copy of everyf.p. group. That is, there exists a universal f.p. group.

Proof. Up to isomorphism, there are countably many f.p. groups. Then their freeproduct A has countably many generators a0, a1, ... with a0 = 1. Let B = A∗〈x, y〉,and consider the subgroups

C = 〈x, any−nxyn : n ∈ N〉H ; D = 〈y, x−nyxn : n ∈ N〉H .Observe that D is a free subgroup of 〈x, y〉, hence of B. It follows, C is a freesubgroup of B and there is an isomorphism C ∼= D via φ : x 7→ y and, for alln ≥ 1, any

−nxyn 7→ x−nyxn. Theorem 2.5a furnishes a group E containing B andsome new letter t in E such that t induces φ by an inner automorphism, that isφ(c) = t−1ct for all c ∈ C. Let F be the subgroup of E generated by x, t. We claimF contains A as a subgroup. For all n ≥ 1, notice

t−1(any−nxyn)t = φ(any

−nxyn) = x−nyxn,

so that there holdsan = tx−nyxny−nx−1yn.

Since y = φ(x) = t−1xt, it follows that an ∈ F for all n ≥ 1. Already thereis nothing to check for a0. We have thus shown our free product A of every f.p.group embeds into a group F with 2 generators. It is a tedious but relativelystraightforward exercise to show F does, in fact, have a recursive presentation. ByTheorem 4.2, F embeds into an f.p. group U . Now, given any f.p. group G, we haveG −→ A −→ F −→ U . We thus obtain a universal f.p. group.

Corollary 5.8. An f.p. group is not universal iff it satisfies a Markov property.

Proof. Fix a Markov property P , so that there exists an f.p. group B not embed-dable into any f.p. group satisfying P . If G is a universal f.p. group, then B embedsinto G. Thus G satisfies no Markov property, lest we contradict our first assertion.Of course, not being universal holds up to isomorphism, there exist non-universalf.p. groups, and the universal f.p. group cannot embed into any non-universal one.That is, not being universal is Markov.

5.3. The Homeomorphism Problem Revisited. Having exhibited some appli-cations of Higman’s theorem to purely group-theoretic decision problems, we nowreturn to the original motivation for this study and consider the general problemof classifying manifolds up to homeomorphism. We have classification theorems formanifolds of dimension up to and including 3. For n ≥ 4, is the set of all closed,connected n-manifolds M with M ≈ N computable for all closed, connected n-manifolds N? To answer this question, we cite Lemma 5.12 in [6].

Lemma 5.9. For all n ≥ 4 and for every finitely presented group H there exists aclosed n-manifold M of dimension greater than 3 with π1(M) ∼= H

22 ALEX BURKA

Theorem 5.10. For all n ≥ 4, the set of simply connected n-manifolds is notcomputable. Thus the homeomorphism problem is not decidable for n ≥ 4.

Proof. The first claim follows immediately from the Adian-Rabin Theorem, the factthat triviality is a Markov property, and Lemma 5.12. The second claim followsfrom the first, and independently follows from Corollary 5.9 and Lemma 5.12.

Acknowledgements

Thanks to Isabella Scott for guidance throughout the program and perceptive feed-back on my paper. Thanks to Shmuel Weinberger for meeting me on a Sundayafternoon to clarify some questions and provide me further directions of study.Finally, thanks to Peter May, whose efforts make this wonderful program possible.

References

[1] Stal Aanderaa. A Proof of Higman’s Embedding Theorem, Using Britton Extensions of Groups.North-Holland Publishing Company. 1973.

[2] Gilbert Baumslag. Topics in Combinatorial Group Theory. Springer Basel AG. 1993.

[3] John L. Britton. The Word Problem. Annals of Mathematics, Second Series, Vol. 77. 1963.[4] S. Barry Cooper. Computability Theory. Chapman and Hall/CRC. 2004.

[5] M. Dehn. Uber unendliche diskontinuierliche gruppen. Annals of Mathematics, First Series,

Vol. 71. 1911.[6] Mathematical Notes. “Amalgamated Products and HNN Extensions.” Link.

[7] David Peifer. Max Dehn and the Origins of Topology and Infinite Group Theory. The Mathe-

matical Association of America. 2015.[8] Joseph Rotman. An Introduction to the Theory of Groups. Springer-Verlag New York, Inc.

1995.[9] John Stillwell. The Word Problem and the Isomorphism Problem for Groups. Bulletin of the

American Mathematical Society, Volume 6, Number 1. 1982.

[10] A. M. Turing. On Computable Numbers, with an Application to the Entscheidungsproblem.Proceedings of the London Mathematical Society. 1936.

[11] Shmuel Weinberger. Computers, Rigidity, and Moduli. Princeton University Press. 2005.