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To adjoin an $n$th root of $g\in G$, let $m$ be the order of a group element $g$ and take a cyclic group $C=⟨a⟩$ of order $nm$ when $m$ is finite, or infinite order when $g$ has infinite order. In the free product $G\ast C$, $⟨a^n⟩$ and $⟨g⟩$ have the same order. The HNN extension with $t^{-1}{a}^{n}t=g$ therefore embeds its base by Britton's lemma, and $h=t^{-1}at$ satisfies $h^n=g$. If $g=1$, take $C$ of order $n$ and identify the two trivial subgroups. This is an existence construction, not an algorithm for finding an unknown element order.

The associated subgroups are the source and target $A,B\le G$ of the isomorphism used to define an HNN extension. They determine its pinches in an HNN extension, its coset transversals, and the possible cancellations of adjacent inverse stable letters.

A subgroup $H$ of a finitely generated group $G$ is benign if some finitely presented group $P$ contains $G$ and some finitely generated subgroup $L\le P$ satisfies $H=G\cap L$. The subgroup $H$ itself need not be finitely generated. A centralizing HNN extension of $P$ along $L$ is finitely presented even when the corresponding intersection subgroup $H$ is infinitely generated.

In the HNN extension $⟨P,t\mid t^{-1}\ell t=\ell (\ell \in L)⟩$, the displayed equivalence follows from Britton's lemma. If $p\in L$, the word collapses by the defining isomorphism; otherwise it has stable letters and no pinch, so it is nonidentity. When $P$ is finitely presented and $L$ is finitely generated, imposing commutation with the finitely many subgroup generators gives a finite presentation. An undecidable membership problem for $L$ then becomes an undecidable word problem for a group in the extension.

Every nontrivial countable torsion-free group embeds in a countable torsion-free group with precisely two conjugacy classes, the identity and all nonidentity elements. At each stage, enumerate all ordered pairs of nonidentity elements of the current group and successively adjoin HNN extensions identifying their infinite cyclic subgroups. Britton's lemma preserves embeddings and torsion in an HNN extension preserves torsion-freeness. Repeat on the resulting countable union, so pairs involving newly introduced elements are eventually included. A final increasing union is countable and torsion-free, and any two nonidentity elements become conjugate at a later stage. No effective enumeration of nonidentity elements is claimed.

A cyclically reduced sequence has no pinch in an HNN extension either internally or across its cyclic junction. If its stable-letter length is positive, every positive power remains reduced, so Britton's lemma gives infinite order. Repeated cyclic conjugation and pinch removal shows that any finite-order element of an HNN extension is conjugate into its base group.

A finite group presentation has finitely many generators and finitely many relators. It defines the quotient of the free group on $X$ by the normal closure of $R$. Free products of finitely presented groups and HNN extensions along explicitly finitely generated associated subgroups have finite presentations obtained by adjoining finitely many generators and relations.

A subgroup is finitely generated when finitely many of its elements generate it as a group. Even inside a finitely generated group, subgroups need not be finitely generated. Finite generation of a witness subgroup, rather than of the original intersection, is what makes centralizing HNN extensions usable for a benign subgroup.

A modular machine can be encoded by HNN extensions of $K={\mathbb{Z}}^2\ast ⟨t⟩$. The kernel of $K\to {\mathbb{Z}}^2$ has a free basis of a group $t(r,s)=x^{-r}{y}^{-s}t{x}^r{y}^s$. Associated-subgroup maps send these basis elements according to the two arithmetic transition types. In the resulting group, membership of $t(r,s)$ in the subgroup generated by $t$ and the instruction stable letters is equivalent to $(r,s)\in H_0(\mathcal{M})$. One more HNN extension centralizes that finitely generated subgroup, converting membership to equality and hence to the word problem for a group.

The two three-generator components on $\{a,b,c\}$ and $\{c,d,a\}$ are iterated HNN extensions of infinite cyclic groups. The subgroup on $\{a,c\}$ is free of rank two in each component by Britton's lemma. Amalgamating over these free subgroups gives the displayed infinite finitely presented group. Its cyclic squaring relations forbid every nontrivial finite quotient by the least-prime multiplicative-order argument.

A group $G$ is a finitely presented group if there is an epimorphism $F(X)\to G$ from a finite-rank free group whose kernel is the normal closure of finitely many words $R$. Equivalently, $G=⟨X\mid R⟩$ with both $X$ and $R$ finite. Normal closure means that every kernel element is a finite product of conjugates of the chosen relators or their inverses.

For independence of finite presentation from finite generators, suppose $G=⟨y_1,\dots ,y_m\mid r_1,\dots ,r_s⟩$, and let $x_1,\dots ,x_k$ be any other finite generating set. Choose words $u_j(X)$ representing $y_j$ and words $v_i(Y)$ representing $x_i$. Consider the finite presentationSending each $x_i$ to its actual element gives a surjection $P\to G$. Conversely, send $y_j$ to $u_j(X)$; the substituted relators make this a homomorphism $G\to P$. Their composition on $G$ fixes every $y_j$, and their composition on $P$ fixes every $x_i$ by the extra relations. They are inverse isomorphisms. Hence finite presentability is independent of the chosen finite generating set.

For subgroups $A,B\le G$ and an isomorphism $\varphi :A\to B$, adopt the conventiona quotient of the free product $G\ast ⟨t⟩$. The generator $t$ is the stable letter of this HNN extension. Fix right coset transversals $R_{+}$ for $B\backslash G$ and $R_{-}$ for $A\backslash G$, each with $1$ representing the subgroup itself. Thus each $g$ has a unique decomposition $g=br$ with $b\in B,r\in R_{+}$, and a unique decomposition $g=ar$ with $a\in A,r\in R_{-}$.

A normal form is a stringwhere $r_i\ne 1$ whenever ${ϵ}_{i+1}=-{ϵ}_i$. The case $n=0$ is just $g_0$. The prohibition removes adjacent inverse stable letters that would cancel. The following permutation proof of HNN normal form uniqueness proves both existence and uniqueness.

Let $\mathcal{N}$ be the set of these formal strings. A base element $g$ acts by left multiplication on the initial coefficient $g_0$; call this permutation $L_g$. To define $L_t$, decompose the initial coefficient as $g_0=br$, with $b\in B,r\in R_{+}$. Unless $r=1$ and the old first stable letter is $t^{-1}$, replace the initial segment byIn the exceptional case, if the old string begins $b{t}^{-1}{r}_1\cdots$, cancel the new $t$ with that $t^{-1}$ and obtain the string beginning ${\varphi }^{-1}(b)r_1\cdots$. These operations use the defining identity $tb={\varphi }^{-1}(b)t$.

Define $L_{t^{-1}}$ similarly: write $g_0=ar$ with $a\in A,r\in R_{-}$, prepend $\varphi (a)t^{-1}r$, and cancel with an old first $t$ exactly when $r=1$. The resulting strings satisfy the normal-form conditions. These two maps are inverses. If the first operation inserts a letter, the reverse operation decomposes its new initial coefficient in the associated subgroup and cancels that letter. If the first operation cancels a letter, the reverse coset decomposition recovers the removed representative and reinserts it; an additional unwanted cancellation is excluded by the original normal-form condition.

The permutations satisfy $L_g{L}_h=L_{gh}$ andIndeed, in the decomposition $g_0=br$, multiplying $g_0$ on the left by $\varphi (a)$ leaves its representative $r$ unchanged and changes $b$ to $\varphi (a)b$. Its image under ${\varphi }^{-1}$ is $a{\varphi }^{-1}(b)$, exactly the coefficient produced by multiplying the normalized result on the left by $a$. The cancellation case has the same representative and the same conclusion. ThereforeAll defining relations of the HNN extension hold in this action, so it induces a homomorphism from $G{\ast }_{\varphi }$ into the permutations of $\mathcal{N}$.

Starting with the empty string, successive left multiplications normalize any word, and every normalization step uses a defining relation. This proves existence. If two normal words represented the same group element, their induced permutations would agree. Applied to the empty string, each normal word gives exactly its own formal string, since its representatives are already chosen and all forbidden cancellations have been excluded. Thus the strings are equal. This proves the normal form theorem for an HNN extension, including the injectivity of the base group $G\to G{\ast }_{\varphi }$.

A reduced sequence in an HNN extension is a wordwith no pinch $t^{-1}at$ for $a\in A$ and no pinch $tb{t}^{-1}$ for $b\in B$. Equivalently, if adjacent exponents are $-1,+1$, their intervening coefficient is outside $A$; if they are $+1,-1$, it is outside $B$.

Normalize such a word from right to left using the procedure above. Inductively its stable-letter length does not change. To see why, suppose the next letter has exponent $\delta$. Moving an associated-subgroup factor through it contributes to its left an element of ${\varphi }^{-1}(B)=A$ when $\delta =1$, or of $\varphi (A)=B$ when $\delta =-1$. If the preceding exponent is $-\delta$, this factor belongs to that preceding letter's pinch subgroup. Multiplying an intervening coefficient outside that subgroup by an element inside it cannot put the product inside the subgroup. Hence no cancellation occurs; when the two exponents agree, cancellation is impossible anyway. Thus a reduced word with $n\ge 1$ has a normal form still containing $n$ stable letters, which cannot be the identity normal form. This is Britton's lemma:

For the final example, write the centralizing extension asIt is generated by $a,b,t$. Suppose it were finitely presented. By the finite-generator independence already proved, the kernel of $F(a,b,t)\to E$ would be normally generated by finitely many words. Each of these kernel group generators is a finite product of conjugates of the displayed defining relators, so only finitely many indices occur in all their expressions. Choose $N$ at least as large as all of them. The full kernel would then equal the normal closure of the first $N$ relators: one inclusion follows from those expressions, and the reverse inclusion follows because all these relators already belong to the kernel.

But the group with only those first $N$ relators is the identity HNN extension over $H_N=⟨w_1,\dots ,w_N⟩$. Since the $w_j$ are a free basis of $H$, $w_{N+1}\notin H_N$. The word $t^{-1}{w}_{N+1}t{w}_{N+1}^{-1}$ is reduced in this partial extension, so Britton's lemma makes it nonidentity there. It is identity in $E$, contradicting equality of the kernels. ThereforeThis is the free-basis instance of the general fact that finite presentation of a centralizing HNN extension detects finite generation of its associated subgroup.

The requested representation is the left coset actionIt is well defined because equal cosets remain equal after multiplication by $g$, and $\rho (g{g}^{\prime })=\rho (g)\rho (g^{\prime })$. It is transitive. For $H=1$ it is the left regular action; for general $H$ it is the associated transitive coset representation.

Its kernel is the normal core of a subgroupIndeed, $g$ fixes every $xH$ precisely when $x^{-1}gx\in H$ for every $x$. Thus $N⊲G$ and $N\le H$. If $[G:H]=i<\infty$, the image lies in the finite symmetric group $S_i$, soIn particular $N$ has finite index in $H$.

For Higman group as an amalgam of iterated HNN extensions, use the cyclic squaring presentationFirst form $L=⟨b,c\mid b^{-1}cb=c^2⟩$, an HNN extension of the infinite cyclic group $⟨c⟩$. Its base embeds, so $c$ has infinite order; $b$ has infinite order as well, since the homomorphism $L\to \mathbb{Z}$ sending $b\mapsto 1,c\mapsto 0$ is onto. Consequently $⟨b⟩$ and $⟨b^2⟩$ are isomorphic infinite cyclic subgroups. Adjoining the stable letter $a$ with $a^{-1}ba=b^2$ givesWe claim $⟨a,c⟩$ is a rank-two free group. No nonzero power of $c$ belongs to $⟨b⟩$: applying the map $b\mapsto 1,c\mapsto 0$ would make that power equal to $1$, contradicting the infinite order of $c$. Thus no nonzero power of $c$ belongs to $⟨b^2⟩$ either. Any nonempty freely reduced word in $a,c$ that contains $a$ is therefore a reduced HNN sequence: between inverse $a$ letters, a nonzero $c$ power cannot create a pinch. Britton's lemma makes it nonidentity. A word containing only $c$ is nonidentity by the base embedding. This proves the claim.

Likewiseis obtained from $⟨d,a\mid d^{-1}ad=a^2⟩$ by adjoining $c$, and its subgroup $⟨c,a⟩$ is free of rank two by the same argument. Form the amalgamated free product $P{\ast }_{⟨a,c⟩}Q$, identifying these two free subgroups generator by generator. Its presentation is exactly that of $\mathcal{H}$. The component embeddings allowed in the question show that it contains $P$, hence is infinite. It has four group generators and four relators, so it is finitely presented.

To show that this Higman group has no nontrivial finite quotient, consider any finite image and the orders of the four generator images. A relation $x^{-1}yx=y^2$ makes $y$ and $y^2$ have the same order, so every generator order is odd. Suppose one is nontrivial, and let $p$ be the smallest prime dividing any of the four orders. Choose a generator $y$ whose order is divisible by $p$, with predecessor $x$ satisfying the displayed relation. Iterating conjugation $|x|$ times givesThe multiplicative order $d$ of $2$ modulo $p$ thus divides $|x|$. Also $d\mid p-1$ and $d>1$, since $2\not\equiv 1(\mathrm{mod}p)$. Any prime divisor of $d$ is smaller than $p$ and divides the order of $x$, contradicting minimality of $p$. Hence every generator image is trivial, so every finite image is trivial. A proper finite-index subgroup would give a nontrivial transitive finite coset image, which is impossible. Therefore $\mathcal{H}$ is infinite and finitely presented, with no proper finite-index subgroups.

A maximal normal subgroup is a proper normal subgroup $M⊲G$ with no normal subgroup strictly between $M$ and $G$. To prove that finitely generated groups have maximal proper normal subgroups, fix a proper normal subgroup $K$ and order the proper normal subgroups containing $K$ by inclusion. For a chain, its union is normal and contains $K$. It is still proper: if the union were all of $G$, each element of a finite generating set would lie in some chain member, and the chain order would put all these finitely many group generators in one member. That member would equal $G$, a contradiction. The Zorn lemma therefore supplies a maximal proper normal subgroup $M$ containing $K$.

Apply this to $\mathcal{H}$ and $K=1$. By the normal-subgroup correspondence, $\mathcal{H}/M$ is nontrivial and simple. It is finitely generated as a quotient of $\mathcal{H}$. It cannot be finite, since $\mathcal{H}$ has no nontrivial finite quotients. Thus

The final assertion for arbitrary proper subgroups is false. Take $G=S_3$ and $H=⟨(12)⟩$. If a normal subgroup contained $H$, it would contain all conjugate transpositions, which generate $S_3$, so it would equal $G$. Hence this proper subgroup lies in no proper normal subgroup, and in particular in no maximal normal subgroup. The maximal-normal existence result requires the starting subgroup itself to be normal.

Suppose $G$ is generated by $d$ elements. An index of a subgroup $n$ gives a transitive action of $G$ on $n$ cosets. After labelling them by $1,\dots ,n$, this is a homomorphism $G\to S_n$, and the subgroup is the stabilizer of one label. There are at most $(n!{)}^d$ homomorphisms, since images of the generators determine the map, and at most $n$ point stabilizers for each map. Thus there are finitely many subgroups of each specified finite index, with the coarse bound $n(n!{)}^d$.

The finite residual isIt is equivalently the intersection of the kernels of all homomorphisms to finite groups. Intersecting all finite-index subgroups gives the same subgroup, since each contains its normal core, a normal subgroup of finite index. A residually finite group is one with $R_G=1$, equivalently every nonidentity element survives in a finite image.

A Hopfian group is one whose surjective endomorphisms are all injective, hence automorphisms. Let $\theta :G\to G$ be surjective, retaining the finite-generation hypothesis. For each $n$, pulling back by $\theta$ sends a normal subgroup of index $n$ to another normal subgroup of the same index. It is injective on this finite set: equality of two preimages implies equality of the subgroups because $\theta$ is surjective. Therefore it is a permutation. For every finite-index normal subgroup $N$, some $m\ge 1$ satisfiesIt follows that $\ker \theta \subseteq \ker {\theta }^m\subseteq N$. Intersecting over $N$ proves finite residual contains surjective endomorphism kernels:If $G$ is also a residually finite group, this forces every surjective endomorphism to have trivial kernel, proving that a finitely generated residually finite group is Hopfian.

For the specified Baumslag-Solitar group, use the convention $[x,y]=xy{x}^{-1}{y}^{-1}$ and writeIn the abelianization, the defining relation gives $2a=3a$, so the image of $a$ is zero, while $t$ generates an unrestricted infinite cyclic group. Thus $G/G^{\prime }\cong \mathbb{Z}$ through the exponent sum of $t$. The kernel is generated by all $a_i$: any word can be rewritten as a product of conjugates of $a$ followed by the final power of $t$. Alternatively $a=[t,a^2]\in G^{\prime }$, so all $a_i\in G^{\prime }$, while killing them leaves a cyclic quotient. Therefore

Define $\theta (a)=a^2$ and $\theta (t)=t$. This respects the defining relation becauseThe image contains $a^2$, $t$, and $t{a}^2{t}^{-1}=a^3$, hence contains $a=a^3(a^2{)}^{-1}$ and is all of $G$. It is not injective. Indeedmaps to $[a^3,a^2]=1$. But it is a reduced word in the HNN extension with associated subgroups $⟨a^2⟩$ and $⟨a^3⟩$. Its potential subwords $t{a}^{\pm 1}{t}^{-1}$ do not have even exponent, and its potential subword $t^{-1}at$ does not have exponent divisible by three. It has no pinch in an HNN extension, so Britton's lemma gives $c\ne 1$. We have explicitly exhibited a surjective, noninjective endomorphism.

It remains to treat the whole second commutator subgroup, not just this one kernel element. SetThese kernels form an increasing sequence of normal subgroups, so $K$ is a normal subgroup. Every iterate is surjective, and the preceding finite residual argument gives $K\subseteq R_G$.

Conjugating the defining relation gives $a_{i+1}^2=a_i^3$ for every integer $i$. Repeated substitution gives, for $j>i$,Also ${\theta }^k(a_i)=a_i^{2^k}$. Thus when $k\ge j-i$,Every pair of the generating elements of $G^{\prime }$ therefore commutes modulo $K$. The image of $G^{\prime }$ in $G/K$ is abelian, which means $G^{\prime \prime }\subseteq K$. This proves the full statement using iterated endomorphism kernels in BS(2,3):

Suppose $G=⟨g_1,\dots ,g_n\mid R⟩$ has a finite group presentation. Work in the free product $B=G\ast F(a,b)$ and putThe $u_i$ freely generate a subgroup of $F(a,b)$. Indeed, in a reduced word in these conjugates, combine adjacent powers with the same index. Expanding the remaining word leaves nonzero powers of $a$ separated by nonzero powers of $b$, so the normal form theorem for a free product makes it nonidentity. This proves the free conjugate bases in a rank-two free group fact we need.

The group homomorphism $B\to F(a,b)$ killing $G$ sends $v_i$ to $a^{-i}b{a}^i$, including $v_0=b$. These are a free conjugate basis by the same argument with $a,b$ interchanged. A nonempty reduced word in the $v_i$ therefore has nontrivial image and cannot be trivial in $B$. Thus $u_i\mapsto v_i$ extends to an isomorphism between the two rank-$(n+1)$ free groups embedded in $B$.

Adjoin one stable letter implementing this isomorphism:We assume Britton's lemma, including its assertion that the base embeds in this HNN extension, and the factor-embedding assertion of the normal form theorem for a free product. Hence $G\hookrightarrow E$. The presentation is finite, and its relations express every generator in $a,t$:Eliminating $b,g_i$ by Tietze transformations therefore gives a two-generator finite presentation with $\boxed{G\hookrightarrow E=⟨a,t⟩}$. This explicitly constructs the two-generator HNN embedding; it does not rely on an arbitrary two-generator embedding preserving finite presentability.

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.

A subgroup $H\le G$ is a benign subgroup when there are an embedding of $G$ into a finitely presented group $E$ and a finitely generated subgroup $L\le E$ with $G\cap L=H$. Finite generation is required of $L$, not of $H$ itself. We use Britton's lemma, the normal form theorem for an amalgamated free product, and, for the final quotient assertion, the Higman embedding theorem.

Choose witnesses $E_H,L_H$ and $E_K,L_K$ for $H,K$. The amalgamated free product $P=E_H{\ast }_G{E}_K$ embeds both witnesses and their common copy of $G$. It is finitely presented: take their two finite presentations and add the finitely many equalities identifying the images of a finite generating set of $G$. These equalities identify all elements of $G$, so they present precisely the amalgam. The original intersections $G\cap L_H=H$ and $G\cap L_K=K$ are preserved by the embeddings.

Now form the finitely presented groupOnly finitely many commutation relators, one for each of the chosen subgroup generators, are needed. The subgroup $Q=⟨G,t,s⟩$ is exactly the multiple HNN extension of $G$ centralizing $H$ and $K$. To prove this claim, take a word in $G,t,s$ reduced for those associated subgroups. It remains reduced in $T$: a $t$-pinch with a coefficient in $G$ occurs precisely when that coefficient lies in $L_H\cap G=H$, and an $s$-pinch precisely when it lies in $K$. Britton's lemma therefore prevents any extra relation, and it also embeds $G$.

Write $G^u=u^{-1}Gu$. The subgroup $L_{\cap }=G^{ts}$ is finitely generated. If $s^{-1}{t}^{-1}gts\in G$, Britton reduction must first pinch $t^{-1}gt$, requiring $g\in H$; its value is then $g$. The remaining $s^{-1}gs$ can represent a base element only if $g\in K$, and then its value is again $g$. Conversely every $g\in H\cap K$ commutes with both letters. Henceso the intersection is benign.

For the join, put $J=⟨H,K⟩$ and $L_{\vee }=⟨G^t,G^s⟩$, again finitely generated. The subgroup generated by $G,G^t,G^s$ in $Q$ isIndeed, expanding conjugate-copy syllables as $t^{-1}gt$ or $s^{-1}gs$ shows that the only possible Britton reductions are exactly the identifications in the indicated amalgamations; a reduced amalgam word therefore has no additional way to become trivial.

We use this elementary consequence of amalgam normal forms: if $C\le D\le A$, then inside $A{\ast }_{C}B$ the subgroup $⟨D,B⟩$ is $D{\ast }_{C}B$ and intersects $A$ exactly in $D$. A reduced expression with a $B\setminus C$ syllable cannot lie in $A$, because multiplying it by an inverse base element still leaves a reduced nontrivial amalgam word. Apply this first to $G{\ast }_H{G}^t$ with $D=J$, obtaining $D_1=⟨J,G^t⟩$ and $D_1\cap G=J$. Since $K\le J\le D_1$, apply it again in $U=(G{\ast }_H{G}^t){\ast }_K{G}^s$ to obtain $⟨D_1,G^s⟩\cap (G{\ast }_H{G}^t)=D_1$.

Each element of $H$ lies in $G^t$, and each element of $K$ lies in $G^s$, since the corresponding stable letter centralizes it. Thus $J\le L_{\vee }$, and $L_{\vee }=⟨D_1,G^s⟩$. Intersecting the previous identities with $G$ givesThis proves the intersection and join of benign subgroups construction with actual finitely generated witnesses inside $T$.

Finally let $N◃G$ be benign, witnessed by $N=G\cap L$ in a finitely presented group $E$. Choose finite generators $X$ for $G$ and express both $X$ and a finite generating set for $L$ as words in $E$. Enumerate every word $w$ on $X$, every word $v$ on those generators of $L$, and every finite proof in the presentation of $E$ that their substituted words are equal. Dovetail these searches and list $w$ when such a proof is found. Exactly the words representing elements of $G\cap L=N$ are listed: membership has a finite expression in the generators of $L$, and every valid equality has a finite relator proof. Since $N$ is normal, these are exactly the null words of $G/N$.

Use this enumerable list as relators on the finite alphabet $X$. Its normal closure is precisely the kernel of $F(X)\to G/N$, so it is a recursive presentation of a group for $G/N$. The Higman embedding theorem now yieldsThis is normal benign subgroup quotient embedding; no word algorithm for $E$ or membership algorithm for $L$ was assumed.

Let $m$ be the order of a group element $g$, allowing $m=\infty$. For finite $m$, take $C=⟨a\mid a^{nm}=1⟩$; for infinite $m$, take $C=⟨a\mid ⟩\cong \mathbb{Z}$. SetThe normal form theorem for a free product embeds both factors, so $G$ is a free factor. The cyclic subgroup $A=⟨a^n⟩$ has order $m$, just as $D=⟨g⟩$ does. Hence $a^{nr}\mapsto g^r$ is a well-defined group isomorphism $A\to D$, and we may form the HNN extensionThe single displayed conjugacy relation implements the full cyclic isomorphism by taking integer powers. Britton's lemma embeds $G_1$ in $H$, so the original $G$ is preserved. Now takeCancellation of successive $t{t}^{-1}$ gives $h^n=t^{-1}{a}^{n}t=g$. This proves the adjoining a root by an HNN extension construction and all required embedding claims. If $g=1$, then $m=1$, $C=C_n$, and $A=D=\{1\}$; the HNN relation is vacuous and the same calculation gives $h^n=1$. The construction also works for $n=1$. It uses the actual order of $g$ as an existence parameter, not an algorithm for determining an unknown order.