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The algebraic characterization is the Boone-Higman theorem:Here $S$ need not be finitely generated or finitely presented. We prove the equivalence using the Higman embedding theorem, Britton's lemma and the normal form theorem for a free product. The original characterization is Theorem I of An algebraic characterization of groups with soluble word problem.

First suppose the embeddings exist. Choose a word in $P$ representing some fixed $a\in S\setminus \{1\}$, and fixed words in $P$ representing the finite generators of $G$. Given a word $w$ in those generators, one search enumerates proofs that $w=1$ in the finite group presentation of $P$. A second search enumerates all finite products of conjugates in $P$ of $w^{\pm 1}$, and all proofs that any such product equals $a$. All these searches are effective: a proof of equality is a finite product of conjugates of defining relators, checked in the free group, and dovetailing handles the countably many candidates.

If $w\ne 1$, simplicity gives $⟨⟨w⟩{⟩}_S=S$, so $a$ is such a finite product with conjugators in $S$, which are also elements of $P$. The second search eventually succeeds. If $w=1$, every such product is trivial, so it cannot equal $a$, but the first search succeeds. Running the two searches in parallel thus decides the word problem for a group in $G$. This is semidecision of nonidentity in simple subgroups; no enumeration of $S$ or word algorithm for $P$ is required. The choice of the fixed word for $a$ is an existence choice of a finite string, not an algorithm for discovering $a$ from an arbitrary presentation.

Conversely suppose the word problem for a group in $G$ is soluble. We construct a recursive simple envelope by cyclic HNN extensions, keeping all effectiveness requirements explicit. Start with $P_0=G\ast ⟨a⟩$, where $a$ has infinite order. Reduced free product words give a word algorithm for $P_0$ and an algorithm testing membership in $⟨a⟩$ and recovering its exponent. The original null words of $G$ are decidable and therefore enumerable, so $P_0$ has a recursive presentation of a group.

Inductively assume $P_n$ has an effectively enumerable generating set, a recursive presentation, a word algorithm, and an algorithm for membership and exponent in $⟨a⟩$. Form $B_n=P_n\ast ⟨z_n⟩$. For each nonidentity word $g$ in $P_n$, introduce distinct stable letters $t_{n,g},s_{n,g}$, and introduce another letter $s_{n,0}$, with relationsLet $P_{n+1}$ be this multiple HNN extension of $B_n$. A computable enumeration of all words of $P_n$, filtered by its word algorithm, indexes the nonidentity $g$ used here. Repeated representatives of the same element cause no difficulty: their stable letters are simply distinct.

Every associated subgroup in these relations is infinite cyclic. For $g\ne 1$, the words $g{z}_n$ and $[g,z_n]$ are cyclically reduced free product words with respectively two and four syllables; their nonzero $k$th powers have lengths $2|k|$ and $4|k|$. Therefore the reduced syllable length of an input bounds the finitely many possible exponents needed to decide membership in each subgroup and find the exponent. Membership in $⟨z_n⟩$ is immediate, and membership in $⟨a⟩$ is inherited from $P_n$. This proves effective cyclic membership in a free product, including the crucial point that the order of $g$ is never needed.

By Britton's lemma, $B_n$ embeds in $P_{n+1}$. To solve its word problem, repeatedly find a pinch in an HNN extension using these membership algorithms, transport the known exponent across its defining isomorphism, and remove two stable letters. With no pinch left, a word retaining stable letters is nontrivial and not in the base; otherwise the base word algorithm applies. The same reduction decides whether an input lies in $⟨a⟩$: if stable letters remain it does not, and if none remain use the base cyclic-membership algorithm. This verifies the inductive effectiveness claims rather than asserting that HNN extensions automatically preserve decidable word problem. Only the finitely many stable letters in the input need inspection.

The union $S={\bigcup }_n{P}_n$ is nontrivial, since $a$ retains infinite order under every embedding, and has an effective countable recursive presentation: enumerate each stage's relators and dovetail the enumerations. We now prove simplicity. Let $D◃S$ contain $g\ne 1$, and take a stage with $g\in P_n$. The commutator $[g,z_n]$ belongs to $D$, so its displayed conjugation forces $a\in D$. Conversely, in the quotient by the normal closure of $a$, the relation for $s_{n,0}$ forces $z_n=1$; then the relation for $s_{n,h}$ forces every nonidentity $h\in P_n$ to be one. Thus the normal closure of $a$ contains each $P_n$. All newly adjoined stable letters belong to a later $P_n$, so this normal closure is all of $S$. Hence $D=S$, proving that $S$ is a simple group containing $G$.

Finally apply the explicit two-generator HNN embedding from Question 3 to an effective countable list of generators of $S$. Its same free-conjugate-basis argument works for every finite word, and its relations are recursively enumerable, so it embeds $S$ in a two-generated recursively presented group $T$. The Higman embedding theorem embeds $T$ in a finitely presented group $P$. Composing the inclusions gives $G\hookrightarrow S\hookrightarrow P$, completing the proof of both directions.

The van Kampen theorem says: if $X=U\cup V$ with $U,V$ open and path connected, $U\cap V$ path connected, and a base point in the intersection, then ${\pi }_1(X)$ is the free product ${\pi }_1(U)\ast {\pi }_1(V)$ modulo the normal subgroup generated by identifying the two images of every element of ${\pi }_1(U\cap V)$. Equivalently it is the pushout of the two inclusion-induced fundamental-group maps.

A Kleinian group is a discrete subgroup of ${\mathrm{PSL}}_2(\mathbb{C})$, acting by Möbius transformations on the Riemann sphere, equivalently by orientation-preserving isometries of hyperbolic space ${\mathbb{H}}^3$.

Consider the classes ofThey generate a discrete subgroup because their matrix representatives have integer entries: a sequence of such matrices converging to $I$ or $-I$ is eventually constant. On the real projective line set $X=\{x:|x|>1\}\cup \{\infty \}$ and $Y=\{x:|x|<1\}$. For every nonzero integer $m$,The last inclusion follows from $|2mx+1|\ge 2|x|-1>|x|$ when $|x|>1$; also $B^m(\infty )=1/(2m)\in Y$. The ping-pong lemma therefore says that the subgroup is the free product $⟨A⟩\ast ⟨B⟩$. Both generators have infinite order, soa free group on two generators. The ping-pong inclusions ensure that no nonempty reduced word can act as the identity, which is the essential freeness argument.

Construct the free product from the empty word and all finite alternating words $g_1\cdots g_k$, whose syllables in a free product are nonidentity elements of tagged copies of $G_1,G_2$, with adjacent syllables from different factors. Multiply by concatenating, multiplying adjacent elements in the same factor, and deleting identities until the word is reduced. The normal form theorem for a free product gives a unique result; reducing three concatenated words gives the same result under either parenthesization, so multiplication is associative. The empty word is the identity, and the inverse reverses the word and inverts each syllable. Each factor embeds as words of length one.

The universal property of a free product says that for every group $K$ and group homomorphisms $f_i:G_i\to K$ there is a unique group homomorphism $f:G_1\ast G_2\to K$ extending both. Explicitly, send a reduced word to the product of its syllable images; reduction preserves this product. This gives existence, while generation by the two factors gives uniqueness.

A standard form of Klein's combination theorem, or the ping-pong lemma, is the following. Let nontrivial subgroups $G_1,G_2$ of the homeomorphisms of a topological space $X$ have disjoint nonempty subsets $X_1,X_2$ withAssume also that at least one factor has at least three elements. Then the generated subgroup is $\boxed{⟨G_1,G_2⟩\cong G_1\ast G_2}$. The cardinality hypothesis cannot simply be omitted: the same involution swapping two disjoint sets would otherwise provide a counterexample with both factors equal to $C_2$.

Here is the ping-pong lemma proof. A reduced word of odd length begins and ends in the same factor, so repeated application of the displayed inclusions sends the other factor's domain into that factor's domain. Disjointness shows that the word is not the identity. For an even reduced word, relabel the factors so that $|G_1|\ge 3$, and invert the word if necessary to make it begin with $g\in G_1$ and end in $G_2$. Choose $s\in G_1\setminus \{1,g^{-1}\}$. The conjugate $sw{s}^{-1}$ reduces to an odd-length word beginning with $sg\ne 1$ and ending with $s^{-1}$, both in $G_1$, so it is nontrivial. Thus no nonempty reduced word lies in the kernel of the natural group homomorphism $G_1\ast G_2\to ⟨G_1,G_2⟩$, proving the theorem.

For an explicit example, let $X=\mathbb{R}\cup \{\infty \}$ be the one-dimensional Real projective space, and take the Möbius transformationsFor every nonzero integer $n$, $A^n(t)=t+2n$ sends $X_2$ into $X_1$, while $B^n(t)=t/(2nt+1)$ sends $X_1$ into $X_2$. Indeed $|2nt+1|>|t|$ when $|t|>1$, and $B^n(\infty )=1/(2n)$. Both transformations have infinite order. Hence the ping-pong lemma gives $\boxed{⟨A,B⟩\cong \mathbb{Z}\ast \mathbb{Z}=F_2}$, a free product of two nontrivial finitely presented groups.

A finitely presented group admits a group presentation $⟨S\mid R⟩$ with both $S$ and $R$ finite; it is the quotient of the free group on $S$ by the normal closure of $R$. Ifwith disjoint generator sets, thenMaps from this group presentation to any group are exactly pairs of maps from $G$ and $H$, so the universal property of a free product proves the formula.

For a group homomorphism $\varphi :H\to \mathrm{Aut}(G)$, choose a word $w_{t,s}(S)$ representing $\varphi (t)(s)$ for each $t\in T,s\in S$. The presentation of a semidirect product isThe presentation maps onto the specified semidirect product. Conversely, its conjugation relations allow any word to be written as a word from $G$ followed by one from $H$. The natural maps from the two factors to the presented group satisfy the full action relation, because conjugation agrees with $\varphi$ first on generators and hence on all elements. They define the reverse group homomorphism $(g,h)\mapsto gh$. The two maps are inverse on every generator, proving the group isomorphism. There are finitely many cross-relations, so the result is again a finitely presented group.

Apply this to the specified permutation action. The presentation isEliminate $y=cx{c}^{-1}$ and $z=c^{2}x{c}^{-2}$. The last cross-relation becomes $c^{3}x{c}^{-3}=x$, already implied by $c^3=1$. ThusBoth factors are nontrivial finitely presented groups, as required.

A useful closed-cover version of the Seifert-van Kampen theorem is the following. Let $X=A\cup B$ with $A,B,A\cap B$ path-connected closed subspaces and basepoint in their intersection. Assume there are open neighbourhoods $U,V$ covering $X$ that deformation retract onto $A,B$, while $U\cap V$ deformation retracts onto $A\cap B$, with the induced inclusions compatible up to based homotopy. ThenThe neighbourhood hypothesis, available for the present finite subcomplex gluing, matters: arbitrary closed covers need not satisfy this theorem.

A Möbius band retracts onto its core circle, with generator $m$; its boundary represents $m^2$. The torus has generators $u,v$ and relation $[u,v]=1$. The attaching circle is the $u$ circle. Reversing the attaching homeomorphism merely replaces $u$ by its inverse and changes no isomorphism type. The closed-cover theorem givesMapping $m\mapsto 0$, $v\mapsto 1$ surjects onto $\mathbb{Z}$, so the group is infinite. Imposing the extra relation $m^2=1$ gives the quotient $C_2\ast \mathbb{Z}$, a nonabelian free product, since alternating reduced words $mv$ and $vm$ differ. A quotient of an abelian group would be abelian, so the fundamental group is nonabelian.

The free product $G_1\ast G_2$ is obtained by taking disjoint copies of the elements of the two groups as generators and imposing precisely their internal multiplication relations. Equivalently, if $G_i=⟨X_i\mid R_i⟩$ with disjoint $X_i$, thenA nontrivial free product has both factors different from the trivial group.

The normal form theorem for a free product says that every element has a unique expressionwhere the empty expression represents the identity. The $g_j$ are called syllables in a free product. Thus nonempty alternating products cannot be the identity, and the natural maps of the factors into the free product are injective.

For existence, multiply adjacent syllables from the same factor and delete identity syllables until the word alternates. Each change uses a defining relation and decreases the number of syllables. To establish uniqueness without assuming that these reductions are confluent, let $\mathcal{W}$ be the set of alternating nonidentity syllable sequences, including the empty sequence. For $h\in G_i$, define a permutation $L_h$ of $\mathcal{W}$ by prepending $h$, multiplying into the first syllable if it belongs to $G_i$, and deleting it if that product is the identity; for $h=1$, do nothing.

The multiplication law inside $G_i$ gives $L_h{L}_k=L_{hk}$ for $h,k\in G_i$, and $L_h^{-1}=L_{h^{-1}}$. This can be checked at the first syllable: if a product deletes it, the next syllable belongs to the other factor and is treated as a new first syllable. Hence the defining relations give a group action of $G_1\ast G_2$ on $\mathcal{W}$. An alternating word $g_1\cdots g_r$ sends the empty sequence to its own syllable sequence. Two such words representing the same group element induce the same permutation, and therefore have identical sequences. This proves the normal form theorem for a free product.

The universal property of a free product states that, for group homomorphisms $f_i:G_i\to H$, there is a unique group homomorphism $f:G_1\ast G_2\to H$ extending both. Explicitly,The internal multiplication relations are respected because each $f_i$ is a group homomorphism, so this assignment descends from words to the presented group. Alternatively, multiplication of two normal forms is concatenation followed by precisely those internal relations, which do not change its value in $H$. Uniqueness holds because the two factors generate the free product.

The base free product for the group encoding of a modular machine isThe letters $x,y$ generate the first factor, and $t$ generates the second. Let $T_K$ be the kernel of the group homomorphism $K\to ⟨x,y⟩\cong {\mathbb{Z}}^2$ that kills $t$. Thenis a free group with this displayed free basis of a group. Indeed every word in $K$ can be rewritten as a product of such conjugates followed by an element of $⟨x,y⟩$, so they generate the kernel. A freely reduced product of these conjugates is nontrivial by the normal form theorem for a free product: after combining adjacent occurrences of the same conjugate, different successive indices give a nonzero intervening $x,y$-syllable. Thus there is no relation among the proposed basis elements.

For clarity, a modular machine of modulus $m>1$ has at most one instruction for each residue pair $(a,b)$, with $0\le a,b<m$ and $0\le c<m^2$. Its right and left transitions are respectivelyIts halting set at a designated terminal configuration $H_0(\mathcal{M})$ consists of the nonnegative pairs whose forward computation reaches $(0,0)$, including $(0,0)$ itself. These formulas explain the exponents in the associated-subgroup maps below.

The factors ${\mathbb{Z}}^2$ and $\mathbb{Z}$ are torsion-free groups, and so is their free product $K$. Indeed every element of a free product is conjugate either into one factor or to an alternating word whose first and last syllables lie in different factors. A word of the latter kind has nonempty reduced powers of every positive length, so has infinite order. A nonidentity element of either factor also has infinite order.

By the finite-order result in part (b), every finite-order element of the multiple HNN extension $K_{\mathcal{M}}$ is conjugate into the torsion-free base $K$, and hence is the identity. Apply the same result once more to $G_{\mathcal{M}}$, whose base is $K_{\mathcal{M}}$. It follows thatThus the failure of a word problem for a group to be decidable does not require torsion.

We construct the output presentation explicitly, using HNN extensions and an amalgamated free product. Throughout, $P$ also denotes the group given by the input finite group presentation, and $X=\{x_1,\dots ,x_n\}$. Every word introduced below is a literal computable word in the displayed generators; testing whether $w=1$ is never part of the construction.

First fix the auxiliary Higman groupWe need two facts: $⟨a_1,a_3⟩$ is a free group of rank two, and the normal closure of $a_1$ is $Q$. The latter follows immediately from the relations: setting $a_1=1$ forces successively $a_2=a_2^2$, $a_3=a_3^2$, and $a_4=a_4^2$, hence all three are $1$.

Here is a normal-form verification of the first fact. The three-generator groupis an amalgamated free product of two Baumslag-Solitar groups $BS(1,2)$ along their infinite cyclic subgroups generated by $a_2$. Their base cyclic groups embed by Britton's lemma, so this amalgamation is legitimate. Nonzero powers of $a_1$ in the first factor lie outside $⟨a_2⟩$, as shown by the homomorphism recording the exponent of its stable letter $a_1$. Nonzero powers of $a_3$ in the second factor also lie outside $⟨a_2⟩$: the stable-letter exponent map first forces a putative equality $a_3^r=a_2^s$ to have $s=0$, and then injectivity of the base cyclic group forces $r=0$. The normal form theorem for an amalgamated free product therefore makes $⟨a_1,a_3⟩$ a rank-two free group in $Q_{123}$.

The same argument applies to $Q_{341}$, formed from the other two cyclic conjugation relations. The full $Q$ is $Q_{123}{\ast }_{⟨a_1,a_3⟩}{Q}_{341}$, so that rank-two free group embeds in $Q$ as well. In particular $a_1$ has infinite order.

Now take the free productand setIntroduce $n+2$ letters $u_0,\dots ,u_{n+1}$, and define the finite presentationFinally impose the Higman relations and identifyThus the algorithm outputsThe symbols $c,C,g_j$ in this display are abbreviations for the explicit words above, not extra generators. Renaming the finitely many new generators to avoid $X$ makes this a uniform effective construction.

An isomorphism-invariant Markov property of finitely presented groups $\rho$ has two finitely presented witnesses: a group $A_{+}$ with $\rho$, and a group $A_{-}$ which cannot embed in any finitely presented group having $\rho$. In symbols,The second condition is an obstruction to embedding, stronger than merely saying that $A_{-}$ itself fails the property.

Let $U$ be a fixed finitely presented group with unsolvable word problem for a group. Form the finitely presented free product $P=U\ast A_{-}$. Given a word $w$ in the generators of $U$, apply part (a) to this $P$ and $w$, and output the finite presentation ofIf $w=1$ in $U$, it is also $1$ in $P$, so $P(w)$ is trivial and $B(w)\cong A_{+}$ has $\rho$. If $w\ne 1$ in $U$, its image stays nontrivial in the free product $P$, and $P$ embeds in $P(w)$. Therefore $A_{-}$ embeds in $B(w)$, which cannot have $\rho$. We have the effective equivalenceAn algorithm recognizing whether an arbitrary finite group presentation has $\rho$ would decide the unsolvable word problem for a group $U$. This contradiction proves the Adian–Rabin theorem: no Markov property of finitely presented groups is algorithmically decidable.

Let $X=Y\cup Z$, where $Y$, $Z$, and $Y\cap Z$ are path-connected subcomplexes containing the basepoint; small open neighbourhoods may be used to reduce to the usual open-cover statement. The Seifert-van Kampen theorem says thatMore explicitly, this is the quotient of the free product ${\pi }_1(Y)\ast {\pi }_1(Z)$ by the normal closure ofEquivalently, ${\pi }_1(X)$ is the pushout of the two inclusion-induced homomorphisms.

Yes. Take $G=F_2$. The free product satisfiesA connected three-sheeted cover of the two-petal rose has fundamental group of rank $1+3(2-1)=4$ by the Nielsen–Schreier formula. Thus $F_4$ is isomorphic to an index-three subgroup of $F_2$. A finite-index subgroup quasi-isometry then gives

The Noncommutative Ruzsa triangle inequality says that nonempty finite subsets $B,C,D$ of a group satisfyThe Ruzsa covering lemma says that if $|CD|\le K|D|$, then some $X\subseteq C$ with $|X|\le K$ satisfies

Now let $A$ be a symmetric subset of a group, so $A^{-1}=A$ and $1\in A$. Apply the triangle inequality with the middle set $A$ to obtain, for $m\ge 4$,The hypothesis therefore gives $|A^m|\le K|A^{m-1}|$. Starting from $|A^3|\le K|A|$ proves

In particular $|A^5|\le K^3|A|$. Apply the covering lemma to $C=A^4$ and $D=A$. There is $X\subseteq A^4$ with $|X|\le K^3$ such thatThe set $A^2$ is symmetric and contains the identity element, so this inclusion is exactly the covering condition showing that

Small doubling alone is insufficient in a noncommutative group. Let $H$ be a finite group, let $G=H\ast ⟨x⟩$ be its free product with an infinite cyclic group, and putThen $A$ is symmetric and $|A^2|\le 5|H|+4=O(|A|)$, while $A^3$ contains the double coset $HxH$. Distinct pairs $(h_1,h_2)\in H^2$ give distinct reduced words $h_{1}x{h}_2$, so $|HxH|=|H{|}^2$. Letting $|H|\to \infty$ proves the small doubling does not control tripling in a noncommutative group phenomenon.

Because the generating sets are disjoint, the free product has the presentationThis is immediate from the universal property of a group presentation: a map out of this presented group is exactly a pair of homomorphisms from $G$ and $H$ to the target group.

Let $H$ be a finite group and form the free product $G=H\ast ⟨x⟩$. Put $A=H\cup \{x\}$. Thenso $|A^2|\le 3|H|+1<3|A|$.

On the other hand, $A^3$ contains the double coset $HxH$. Reduced-word uniqueness in the free product makes the mapinjective, so $|A^3|\ge |H{|}^2$. Taking finite groups $H$ of unbounded order keeps the doubling constant below three while $|A^3|/|A|$ tends to infinity. This realizes the small doubling does not control tripling in a noncommutative group phenomenon.