HNN extension

Given isomorphic subgroups $C,D\le A$ and an isomorphism $\varphi :C\to D$, the HNN extension isIts stable letter $t$ records the loop edge in the associated graph of groups.

Every finite-order element of an HNN extension is conjugate to a finite-order element of its base. Choose a conjugate minimizing the number of stable letters. If this number is positive, no pinch is possible either internally or across the end-to-start join, since a boundary pinch would allow a conjugation shortening it. Every positive power is then reduced with positive stable-letter length, and Britton's lemma makes it nonidentity. In particular, an HNN extension of a torsion-free group is torsion-free.

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.

For a finitely presented base group $G$ and subgroup $H$, the extension $⟨G,t\mid [t,h]=1(h\in H)⟩$ is finitely presented exactly when $H$ is finitely generated. Sufficiency follows by imposing commutation with finitely many subgroup generators. For necessity, any finite set normally generating the defining kernel uses only finitely many centralizing relators. If their subgroup $H_0$ is proper in $H$, Britton's lemma makes $t^{-1}ht{h}^{-1}$ nonidentity in the partial extension for $h\in H\setminus H_0$, contradicting the claimed kernel equality.

// Target: algebra.bigb

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.

Fix right coset transversals $R_{+},R_{-}$ for $A,B$, containing $1$. In the convention $ta{t}^{-1}=\varphi (a)$, every element has a unique normal form $g_0{t}^{{ϵ}_1}{r}_1\cdots t^{{ϵ}_k}{r}_k$, with $r_i\in R_{{ϵ}_i}$ and $r_i\ne 1$ whenever ${ϵ}_{i+1}=-{ϵ}_i$. A permutation action on the set of these formal sequences proves uniqueness and embeds the base group. Transporting subgroup coefficients across stable letters converts general words to these normal forms.

With convention $t^{-1}at=\varphi (a)$, choose representatives of the right cosets of $A$ and $B$ in the base group, with identity representing each subgroup. Normal strings have the form $g_0{t}^{{ϵ}_1}{r}_1\cdots t^{{ϵ}_n}{r}_n$, where $r_i$ represents a coset of $B$ for ${ϵ}_i=1$ and of $A$ for ${ϵ}_i=-1$, and $r_i\ne 1$ when adjacent exponents are opposite. Left multiplication by base elements or stable letters defines permutations of these strings. They satisfy the HNN relations. Applying the induced action to the empty string distinguishes different normal strings, proving uniqueness without presupposing Britton's lemma.

// Target: geometry-and-topology.bigb

A reduced sequence is a word $g_0{t}^{{ϵ}_1}{g}_1\cdots t^{{ϵ}_k}{g}_k$ with no pinch in an HNN extension. If $k>0$, Britton's lemma makes it nontrivial and prevents it from representing a base-group element. Reduced sequences need not be unique; uniqueness requires the fixed coset representatives in the normal form theorem for an HNN extension.

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.

In $⟨G,t\mid ta{t}^{-1}=\varphi (a),a\in A⟩$, a pinch is a subword $ta{t}^{-1}$ with $a\in A$, or $t^{-1}bt$ with $b\in B=\varphi (A)$. It can be replaced by a base-group element, removing two stable letters. Britton's lemma asserts that every identity word with stable letters must have a pinch after base-group multiplication is performed.

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 stable letter is the new generator of an HNN extension which implements an isomorphism between two specified subgroups by conjugation. In the convention $ta{t}^{-1}=\varphi (a)$, it moves elements of the source subgroup $A$ to the target subgroup $B$; using $t^{-1}at=\varphi (a)$ reverses the naming convention.

In an HNN extension, every word containing a stable letter and having no pinch $t^{-1}ct$ or $td{t}^{-1}$ is nonidentity. In particular, the natural map from the base group to its HNN extension is injective.

The Baumslag-Solitar group is

Represent $y$ by translation $t\mapsto t+1$ and $x$ by $t\mapsto t/3$; the relator holds and $y$ has infinite order. Base-three digits $d_j\in \{0,1,2\}$ give words recursively by $W_0=1$, $W_{j+1}=x^{-1}{W}_{j}x{y}^{d_j}$. These have length at most $4k$ and represent all $y^m$, $0\le m<3^k$. Thus the growth function of a finitely generated group obeys $\beta (4k)\ge 3^k$. Counting freely reduced words in four letters gives $\beta (n)\le 2\cdot 3^n-1$. The growth is therefore exponential up to coarse equivalence.

For $G=⟨a,t\mid t{a}^2{t}^{-1}=a^3⟩$, the surjective endomorphism $\theta (a)=a^2$, $\theta (t)=t$ has nontrivial kernel by Britton's lemma. The commutator subgroup is generated by $a_i=t^{i}a{t}^{-i}$ for $i\in \mathbb{Z}$. For $j>i$, its defining relations give $a_j^{2^{j-i}}=a_i^{3^{j-i}}$. Hence ${\theta }^k([a_i,a_j])=1$ whenever $k\ge j-i$. The union of the increasing kernels is normal, and the images of all $a_i$ commute in its quotient, proving the first inclusion. The second follows from finite residual contains surjective endomorphism kernels applied to every iterate.

In the finite group presentation $⟨a,t\mid ta{t}^{-1}=a^2⟩$, let $a$ act by $x\mapsto x+1$ and $t$ by $x\mapsto 2x$. The conjugates $a_i=t^{-i}a{t}^i$ satisfy $a_{i+1}^2=a_i$, so their increasing cyclic subgroups give the additive group $\mathbb{Z}[1/2]$. Every element has the form $a_i^m{t}^k$, and the affine action proves that this form can be the identity only if $k=m=0$. Thus the displayed semidirect product identification holds. This is a finitely presented group with a series having torsion-free abelian factors, but it is not a poly-infinite cyclic group, since $\mathbb{Z}[1/2]$ is not finitely generated.