Faithful inner-homogeneous $R$-representations via HNN extensions with cardinal bound $\max\{\aleph_0,\allowbreak|R|,\allowbreak|G|,\allowbreak|V|\}$

Proof. For $\varepsilon\in\{+1,-1\}$ put

Choose sets $T_\varepsilon$ of representatives such that every $g\in G$ has a unique expression $g=rc$, with $r\in T_\varepsilon$ and $c\in C_\varepsilon$, and such that $1\in T_\varepsilon$ represents $C_\varepsilon$. Consider formal strings

subject to

The case $m=0$ is the string $g$.

Write $P_m=r_1t^{\varepsilon_1}\cdots r_mt^{\varepsilon_m}$, with $P_0=1$. For $g=rc$, $r\in T_\varepsilon$, $c\in C_\varepsilon$, define

Both outputs satisfy (12). Since $\psi_\varepsilon(c)\in C_{-\varepsilon}$, reversing a noncancelling step deletes the added block and restores $r\psi_{-\varepsilon}\psi_\varepsilon(c)=rc$. After a cancelling step, $r_m\psi_\varepsilon(c)$ is already in $T_{-\varepsilon}C_{-\varepsilon}$ form; reversal restores the deleted block. A second cancellation would violate (12). Hence the two stable-letter operations are inverse.

For $a\in A$, write the tail as $rc$, $c\in B$. Appending $ta$ and $\theta(a)t$ uses the same representative $r$ and the same case above; in either case the new tail is $\theta^{-1}(c)a$, preceded by $r_m$ in the cancelling case. Thus

Together with $(w\cdot b)\cdot b'=w\cdot(bb')$, this defines a right $H$-action on the normal strings.

Every operation is valid in $H$, by

Reading a normal string from $1$ returns that string. Equal elements of $H$ therefore have equal normal strings, so $G\to H$ is injective. In particular, a word

with no segment $tat^{-1}$, $a\in A$, or $t^{-1}bt$, $b\in B$, lies outside $G$. Before a first cancellation, the coefficient carried past $t^{\varepsilon_i}$ lies in $C_{-\varepsilon_i}$. For opposite successive signs, multiplication by this coefficient preserves membership in $C_{-\varepsilon_i}$; cancellation would require one of the excluded segments. The normal string consequently retains all $m$ stable letters.

The endpoints in (9) are well-defined, because

They are distinct, since $t\notin G$. The graph is connected: its edges permit multiplication by a stable letter in either direction, while $G$ fixes the base vertex, and $G\cup\{t\}$ generates $H$.

More explicitly, a step from $wG$ can be written $wG\to wgt^\varepsilon G$ with $g\in G$. For $\varepsilon=-1$ the edge is $wgA$; for $\varepsilon=+1$ it is $wgtA$ traversed backwards. Two consecutive steps containing $tat^{-1}$, $a\in A$, traverse the same edge in opposite directions. The same holds for $t^{-1}bt$, $b\in B$, using $bt=t\theta^{-1}(b)$. A nonbacktracking path consequently gives a word of the form (16) with none of the excluded segments. A positive length closed path based at $G$ would give such a word lying in $G$, contradicting the reduced-word assertion. By translating the base vertex, the same holds at every vertex. The graph is a tree. ◻

Proof. For $0\ne x\in\bigoplus_eN_e$, put $F=\operatorname{supp}x$. A component of the finite forest $F$ containing an edge has two leaves, the endpoints of a longest simple path. At either leaf $w$, let $e$ be its unique incident edge in $F$. Then

The last inequality uses $x_e\ne0$ and the injectivity of the endpoint maps. Thus $|\operatorname{supp}(\partial x)|\ge2$, giving both $\partial x\ne0$ and $\partial x\notin\iota_v(M_v)$ for every $v$. The induced map $M_v\to\operatorname{coker}\partial$ is therefore injective. ◻

Proof. Take $H$ from (8). By Lemma 2 we may identify $G$ with its image in $H$. Form the induced modules

Define an $R[H]$-linear map $\delta:E\to M$ by

To check balancing, let $a\in A$. Then

The formula is $R$-linear and satisfies $\delta(kh\otimes u)=k\delta(h\otimes u)$, so it defines an $R[H]$-linear map.

For each $x\in H/G$ choose a representative $r_x$, with $r_G=1$. The right $R[G]$-module decomposition $R[H]=\bigoplus_x r_xR[G]$ gives

where each $V_x$ is a copy of $V$ and each $U_y$ is a copy of $U$. These are decompositions of $R$-modules; no basis of $V$ or $U$ is assumed.

Choose a representative $s_y$ for each $y\in H/A$. Write

where $x=s_yG$ and $z=s_yt^{-1}G$ are the endpoints of $y$. The map from the summand $U_y$ in (27) has exactly the two components

Both endpoint maps are injective: the inclusions $U,U'\le V$ are injective, $\varphi$ is an isomorphism, and the actions of $g_0,g_1$ are invertible. By Lemma 2, the indexing graph in (27) is a tree. Lemma 3 therefore gives

Set

Equivalently, let $D=\bigoplus_{h\in H}V$ with summand maps $e_h$, and let $C\leq D$ be the $R$-submodule generated by

Then $\widehat V_0\cong D/C$. Indeed, quotienting by the first family gives $M$ through $e_h(v)\mapsto h\otimes v$; the inverse is the balanced map $h\otimes v\mapsto[e_h(v)]$. Under this identification, the second family is the negative of $\delta(ht\otimes u)$ and spans $\operatorname{im}\delta$. The submodule $\operatorname{im}\delta$ is $H$-invariant, so $\widehat V_0$ is an $R[H]$-module. Equation (30) makes $j_0$ injective. Equivariance follows from

For the stable letter,

The conjugation equation comes from the presentation of $H$.

Put

For the diagonal action $\sigma$, the regular basis gives

Thus $\ker\sigma=1$, and (34) still holds for $j$.

For $\kappa$ as in (23), finite words and finite sums give

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Proof. On the two disjoint unions, put

Directedness and compatibility make these equivalence relations. Set $G=(\coprod_iG_i)/{\sim}$ and $V=(\coprod_iV_i)/{\sim}$. For $k\ge i,j$, define

and

Passing to a common upper stage proves well-definedness and the representation identities. Injectivity follows from

for some $k\ge i$, and the analogous calculation for $b_{ik}$. Finite sets have representatives in one stage. A compatible family $(f_i)$ factors uniquely by $[i,x]\mapsto f_i(x)$ on each sort.

If $a\ne b$ in $G_i$ and $\ker\rho_i=1$, choose $v\in V_i$ with $av\ne bv$. Then

which proves $\ker\rho=1$.

Each quotient has cardinality at most that of the corresponding disjoint union. Under (42),

These inequalities use only $\kappa\cdot\kappa=\kappa$ for infinite $\kappa$. ◻

Proof. Apply $a$ and $b$ to the two implementation equations. The required identities follow from

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Proof. By (2),

Fix $S=(\langle a_1,\ldots,a_m\rangle,\sum_{j=1}^nR[A]u_j)$. An isomorphism with domain $S$ is determined by

the group images determine $\theta$, and for finite $F_j\subseteq A$,

The codomain is $(\theta(A),\varphi(U))$. There are at most $\kappa^{m+n}\le\kappa$ choices for (52), hence $|\mathcal D(X)|\le\kappa\cdot\kappa=\kappa$. ◻

Proof. By Lemma 7, write $\mathcal D(X)=\{d_\alpha:\alpha<\lambda\}$, $|\lambda|\le\kappa$. Construct compatible embeddings by transfinite recursion:

At a successor, transport $d_\alpha$ along $X\hookrightarrow X_\alpha$ and realize it by Theorem 4; at a limit, use Lemma 5. Induction gives

since $|\beta|\kappa\le\kappa$ at every limit $\beta\le\lambda$. Lemma 6 preserves all previous realizations, so $X^+=X_\lambda$ suffices; for $\lambda=0$, take $X^+=X$. ◻

Proof of Theorem 1. Let $\kappa$ be as in (6). The representation

with diagonal action contains $(G,V)$ by $v\mapsto(v,0)$ and satisfies $\ker\rho_{X_0}=1$ and $|G_{X_0}|,|V_{X_0}|\le\kappa$. Apply Proposition 8 successively to obtain

where $\mathcal D(X_n)$ is realized in $X_{n+1}$. Lemma 5 gives an extension $(\Gamma,W,\sigma)$ with

Identify each stage with its image in this direct limit. These images are nested, so

Let $(S,T,(\theta,\varphi))\in\mathcal D(\Gamma,W)$, with $S=(A,U)$ and $T=(B,U')$. Choose finite group generating sets $S_A,S_B$ and finite module generating sets $T_U,T_{U'}$ over $R[A],R[B]$, respectively. By Lemma 5, some $X_n$ contains all four finite sets. Closure of that stage under its operations gives

The stage actions are the restrictions of the limit action, so $(S,T,(\theta,\varphi))\in\mathcal D(X_n)$. Choose the implementing $t\in G_{n+1}$. By Lemma 6, (7) holds in $(\Gamma,W)$. The automorphism $(c_t,L_t)$, with inverse $(c_{t^{-1}},L_{t^{-1}})$, extends the given isomorphism. ◻

Proof. The cardinal $\kappa$ in (6) is $\aleph_0$. ◻

Proof. Apply Theorem 1 with $R=K$. ◻