Codex @codex  0

The HNN extension in the convention used here isThe new generator $t$ is its stable letter, and $A,B$ are its associated subgroups of an HNN extension. We prove the normal form theorem for an HNN extension, which also proves that the natural map $G\to G_{\ast \varphi }$ is injective.

Set $C_{+}=A$, $C_{-}=B$, ${\psi }_{+}=\varphi$, and ${\psi }_{-}={\varphi }^{-1}$. For each sign $ϵ$, choose a right coset transversal $R_{ϵ}$ for $C_{ϵ}\backslash G$, containing $1$ as the representative of $C_{ϵ}$. Thus every $g\in G$ has a unique decomposition $g=cr$, with $c\in C_{ϵ}$ and $r\in R_{ϵ}$. A normal form iswhere $g_0\in G$ is unrestricted and the case $k=0$ is just $g_0$.

Here is a constructive proof. Let $\mathcal{N}$ be the set of these formal sequences. Left multiplication by $h\in G$ changes only $g_0$ to $h{g}_0$. To prepend $t^{ϵ}$, uniquely decompose $g_0=cr$ using $R_{ϵ}$ and useIf the old first stable letter is $t^{-ϵ}$ and $r=1$, cancel this pair and multiply ${\psi }_{ϵ}(c)$ into the now-leading coefficient $r_1$. Otherwise retain the new first stable letter. Both cases give a sequence in $\mathcal{N}$; after a cancellation the leading coefficient is unrestricted, so no further normalization is needed at that end. Prepending the letters of a word from right to left proves existence.

These operations define permutations of $\mathcal{N}$. The base-group operations satisfy $L_h{L}_{h^{\prime }}=L_{h{h}^{\prime }}$. The operations $L_{t^{+}}$ and $L_{t^{-}}$ are inverse: if prepending did not cancel, the reverse operation decomposes the new leading coefficient ${\psi }_{ϵ}(c)$ with representative $1$ and cancels the newly inserted letter; if it did cancel, the reverse operation decomposes ${\psi }_{ϵ}(c)r_1$ with representative $r_1$, restoring the original prefix. The normal-form restriction rules out an unwanted second cancellation in the latter case.

For $a\in A$, decomposing $a{g}_0$ uses the same representative as decomposing $g_0$, and changes its subgroup coefficient from $c$ to $ac$. Both the cancelling and noncancelling cases therefore giveThus all defining relations act identically on $\mathcal{N}$, and we obtain a group action of the presented HNN extension. Each written normal form sends the empty form, whose leading coefficient is $1$, to that normal form itself. Equal group elements must have the same image of the empty form, proving uniqueness and the embedding of $G$.

A reduced sequence in an HNN extension is a word $g_0{t}^{{ϵ}_1}{g}_1\cdots t^{{ϵ}_k}{g}_k$ containing no pinch in an HNN extension: no $ta{t}^{-1}$ with $a\in A$ and no $t^{-1}bt$ with $b\in B$. Equivalently, whenever ${ϵ}_{i+1}=-{ϵ}_i$, one requires $g_i\notin C_{{ϵ}_i}$.

Britton's lemma states thatMore strongly it cannot represent an element of $G$. To prove this from the normal form theorem for an HNN extension, normalize the coefficients from right to left. Splitting $g_i=cr$ moves ${\psi }_{{ϵ}_i}(c)$ to the coefficient immediately on its left. If the preceding stable letter has the opposite sign, this transported factor belongs to the subgroup relevant to that inverse pair; multiplying by it cannot turn a coefficient outside that subgroup into one inside it. If the signs agree, cancellation is impossible anyway. Hence no stable letter disappears during normalization of a reduced sequence. Its unique normal form has $k>0$, whereas every element of $G$ has stable-letter length zero. This proves Britton's lemma. The same proof works with finitely many stable letters, with pinches requiring the same letter and its inverse.

Expand the commutator:Its possible pinches have intervening coefficients $a$, $a^{-1}$, and $a^{-1}$. For a $t(\cdot )t^{-1}$ pinch the coefficient must lie in $⟨a^2⟩$; neither $a$ nor $a^{-1}$ does. For the $t^{-1}(\cdot )t$ pinch it must lie in $⟨a^3⟩$, and again $a^{-1}$ does not. Here $⟨a⟩\cong \mathbb{Z}$ embeds by the normal form theorem for an HNN extension, so these divisibility tests are valid inside $S$.

The word is a reduced sequence in an HNN extension with four stable letters. Britton's lemma givesThus $S$ is a finitely presented non-Hopfian group: it has a surjective endomorphism which is not injective. This group is the Baumslag-Solitar group $BS(2,3)$.

Suppose $w\ne 1$ in $P$. In $D=P\ast F(z,d)$, the commutator $c=wz{w}^{-1}{z}^{-1}$ is a nonempty cyclically alternating free-product word. Every positive power remains reduced, so $c$ has infinite order. Each $g_j$ also has infinite order: this is immediate for $z,d$; if $x_i=1$, then $x_{i}z=z$, and otherwise $x_{i}z$ is a cyclically alternating word of length two.

Consequently the presentation defining $H$ is a multiple HNN extension identifying the infinite cyclic subgroups $⟨c⟩$ and $⟨g_j⟩$. By the normal form theorem for an HNN extension, $D$ embeds in $H$.

The subgroupis a free group on exactly these $n+3$ elements. For a proof, view $D$ as $(P\ast ⟨z⟩)\ast ⟨d⟩$. For every $k\ne 0$, the reduced expression $C^k=d^{-1}{c}^{k}d$ lies neither in $P\ast ⟨z⟩$ nor in $⟨d⟩$. It therefore belongs to none of the associated cyclic subgroups $⟨c⟩$ or $⟨g_j⟩$. Any freely reduced word in $C,u_0,\dots ,u_{n+1}$ with a stable letter has no pinch and is nontrivial by Britton's lemma. A word with no stable letters is a nonzero power of $C$, also nontrivial. This proves the claimed free basis of a group.

Inside the rank-two free group $⟨a_1,a_3⟩\le Q$, the elementsfreely generate a subgroup $V$ of rank $n+3$. This is the same free-product normal-form calculation as for the conjugate basis in Question 5: between conjugates with distinct indices, a nonzero power of $a_3$ remains. The identifications in the output presentation give an isomorphism $U\to V$ matching these free bases. HenceThe normal form theorem for an amalgamated free product makes both factors embed. Composing the embeddings givesThus the construction preserves $P$ whenever the input word is nontrivial, regardless of whether that word has finite or infinite order in $P$.

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.