Proof. By [1], the map \(\nu\) is a local valuation on \(A\).

\((1)\Rightarrow(2)\). Suppose that \(E^0=\{\mathtt{v}\}\). Then \(\mathtt{v}=1\), and hence \(A=A\mathtt{v}=\mathtt{v}A\). It follows from Definition 5(3) that \(\nu(ab)=\nu(a)+\nu(b)\) for all \(a,b\in A\).

\((2)\Rightarrow(3)\). Suppose that \(\nu\) is a valuation. If \(a,b\in A\) are nonzero, then \(\nu(a),\nu(b)\in\mathbb{N}_0\), and so \(\nu(ab)=\nu(a)+\nu(b)\neq -\infty\). Hence \(ab\neq 0\). Thus \(A\) is a domain.

\((3)\Rightarrow(1)\). This is precisely [1] in the case of LV-graphs: since \(K\) is a field and \((E,\omega)\) is an LV-graph, \(L_K(E,\omega)\) is a domain only when \((E,\omega)\) is an LV-rose.

\((1)\Rightarrow(4)\). Write \(E^0=\{\mathtt{v}\}\). Then \(A\) is unital with identity \(\mathtt{v}\). Let \(u\in A^\times\), and let \(u^{-1}\) be its inverse. By \((2)\), we have \[0=\nu(\mathtt{v})=\nu(uu^{-1})=\nu(u)+\nu(u^{-1}).\] Since \(u\) and \(u^{-1}\) are nonzero, both \(\nu(u)\) and \(\nu(u^{-1})\) lie in \(\mathbb{N}_0\). Thus \(\nu(u)=\nu(u^{-1})=0\). By the definition of \(\nu\), this means that \(\operatorname{supp}(u)\subseteq E^0\) and \(\operatorname{supp}(u^{-1})\subseteq E^0\). Hence \(u=r\mathtt{v}\) and \(u^{-1}=s\mathtt{v}\) for some nonzero \(r,s\in K\). From \(uu^{-1}=u^{-1}u=\mathtt{v}\), we get \(rs=sr=1\). Thus \(r\in K^\times\), and \(u\in K^\times \mathtt{v}\). The reverse inclusion is immediate, since \((r\mathtt{v})^{-1}=r^{-1}\mathtt{v}\) for \(r\in K^\times\). Therefore \(A^\times=K^\times \mathtt{v}=K^\times 1\).

\((4)\Rightarrow(1)\). Suppose by contradiction that \((E,\omega)\) is not an LV-rose. If \(A\) is not unital, then (4) fails. Thus assume that \(A\) is unital. Then \(E^0\) is finite and \(|E^0|>1\). Since \(E\) is connected, there is a structured edge \(\mathtt{e}\) with \(s(\mathtt{e})\neq r(\mathtt{e})\). Choose \(1\leq i\leq \omega(\mathtt{e})\) and put \(p=\mathtt{e}_i\). Then \(p\neq 0\) and \(p=s(\mathtt{e})p r(\mathtt{e})\). Since \(s(\mathtt{e})r(\mathtt{e})=0\), we have \(p^2=0\). Therefore \(1+p\) is an invertible element, with inverse \(1-p\).

This invertible element is not a scalar multiple of \(1\). Indeed, if \(1+p=r1\) for some \(r\in K^\times\), then multiplying on the left by \(s(\mathtt{e})\) and on the right by \(r(\mathtt{e})\) gives \(p=0\), a contradiction. Hence \(A^\times\neq K^\times 1\), so (4) fails. Thus the implication \((4)\Rightarrow(1)\) follows. ◻

Proof. Write \[\omega(s^{-1}(\mathtt{v})) =\{\omega_1<\cdots<\omega_k\},\] and, for \(1\leq j\leq k\), let \[n_j=\bigl|\{\mathtt{e}\in E^{\mathrm{st}} \mid \omega(\mathtt{e})=\omega_j\}\bigr|.\] Since \((E,\omega)\) is a rose, the relations of \(M_{(E,\omega)}\) given in (1 ) take the form \[\label{eq:rose-monoid-relation} q^{\mathtt{v}}_{j-1} +(\omega_j-\omega_{j-1})\mathtt{v} = q^{\mathtt{v}}_j+n_j\mathtt{v}, \qquad 1\leq j\leq k,\tag{2}\] where \(q^{\mathtt{v}}_0=q^{\mathtt{v}}_k=0\) and \(\omega_0=0\).

We first consider the case \(k=1\). If \(\omega_1=1\), then all weights are trivial, so \(L_K(E,\omega)\) is the ordinary Leavitt path algebra of the rose with \(n_1\) petals. Since \((E,\omega)\) is not \(R_1\), we have \(n_1\geq2\). Thus the conclusion holds by taking \(G\) to be this ordinary rose.

Suppose now that \(k=1\) and \(\omega_1=N\geq2\). Since \((E,\omega)\) is not an LV-rose, there is only one structured loop, say \(\mathtt{e}\), and \(\omega(\mathtt{e})=N\). Let \(G=R_N\) be the ordinary rose with loops \(c_1,\ldots,c_N\). Then the assignments \[c_i\longmapsto \mathtt{e}_i^*, \qquad c_i^*\longmapsto \mathtt{e}_i, \qquad v_G\longmapsto\mathtt{v},\] gives \[L_K(G)\cong L_K(E,\omega).\] Since \(N\geq2\), it follows that \(L_K(G)\) is noncommutative.

It remains to consider \(k\geq2\). We first observe that \[q^{\mathtt{v}}_j\neq0 \qquad\text{for }1\leq j\leq k-1.\] Let \(\mathcal{B}=\{0,\infty\}\) be the commutative monoid with \(0+\infty=\infty+0=\infty\) and \(\infty+\infty=\infty\). Assign \[\mathtt{v}\longmapsto\infty \quad\text{and}\quad q^{\mathtt{v}}_j\longmapsto\infty \quad (1\leq j\leq k-1).\] Since \(\omega_j-\omega_{j-1}>0\) and \(n_j\geq1\), both sides of every relation 2 are sent to \(\infty\). Thus these assignments induce a monoid homomorphism \(M_{(E,\omega)}\longrightarrow\mathcal{B}\). In particular, we have that \(q^{\mathtt{v}}_j\ne 0\) for all \(1\leq j\leq k-1\). We now obtain a relation \[\label{eq:v-mv-q} \mathtt{v}=m\mathtt{v}+\mathtt{q}\quad\text{with } m\geq1\text{ and }\mathtt{q}\neq0.\tag{3}\] If \(\omega_1=1\), relation 2 for \(j=1\) has the form \(\mathtt{v}=n_1\mathtt{v}+q^{\mathtt{v}}_1\), so we take \(m=n_1\) and \(\mathtt{q}=q^{\mathtt{v}}_1\). If \(\omega_1\geq2\), then \((E,\omega)\) fails to be an LV-rose only because there is a unique structured edge of maximal weight. Thus \(n_k=1\), and the relation for \(j=k\) becomes \[q^{\mathtt{v}}_{k-1} +(\omega_k-\omega_{k-1})\mathtt{v} =\mathtt{v}.\] In this case take \[m=\omega_k-\omega_{k-1} \quad\text{and}\quad \mathtt{q}=q^{\mathtt{v}}_{k-1}.\] Let \(G\) be the following ordinary graph:

When \(m=1\), the list \(z_1,\ldots,z_{m-1}\) is empty. Its graph monoid has the presentation \[M_G= \left\langle z_1,\ldots,z_{m-1},v_G,q_G \;\middle|\; z_i=v_G,\quad v_G=mv_G+q_G \right\rangle .\] Under the identification \(V(L_K(G))\cong M_G\), the distinguished element is \[\relax[L_K(G)]=z_1+\cdots+z_{m-1}+v_G+q_G=mv_G+q_G=v_G.\] Define \[\overline{\phi}:M_G\longrightarrow M_{(E,\omega)}\] by \[\overline{\phi}(z_i)=\mathtt{v}, \qquad \overline{\phi}(v_G)=\mathtt{v}, \qquad \overline{\phi}(q_G)=\mathtt{q}.\] Relation 3 shows that \(\overline{\phi}\) is well-defined. Moreover, \[\overline{\phi}([L_K(G)]) =\mathtt{v} =[L_K(E,\omega)].\] Thus \(\overline{\phi}\) is a homomorphism of monoids with distinguished element. By Theorem 2, there is a unital \(K\)-algebra homomorphism \[\phi:L_K(G)\longrightarrow L_K(E,\omega)\] which induces \(\overline{\phi}\) on \(V\)-monoids. The graph \(G\) satisfies Condition (L), thus the Cuntz–Krieger uniqueness theorem therefore implies that \(\phi\) is injective; see [9]. Hence \(L_K(G)\) embeds unitally into \(L_K(E,\omega)\). Also it is clear that \(L_K(G)\) is noncommutative. This completes the proof. ◻

Proof. It follows from Theorem 3 that there exists a graph \(F\) such that \(L_K(F)\) is non-commutative and embedded as a subring into \(L_K(E,\omega)\). By [5], we conclude that \(L_K(F)^\times\) contain a non-cyclic free subgroup, so does \(L_K(E,\omega)^\times\). ◻

Proof. Put \(\mathtt{u}=s(\mathtt{e})\) and \(\mathtt{v}=r(\mathtt{e})\). We begin with the smallest possible lift of the edge \(\mathtt{e}_i\). Let \(F_0\) be the directed graph with two vertices \(u,v\) and one edge \(e_i\) from \(u\) to \(v\). Define \[\phi_0^0(u)=\mathtt{u},\qquad \phi_0^0(v)=\mathtt{v},\quad\text{and}\quad \phi_0^1(e_i)=\mathtt{e}_i.\] Then \(\phi_0:F_0\to \widehat E\) is a graph homomorphism.

The graph \(F_0\) need not satisfy the two conditions in Definition 2. We will enlarge it step by step. At each stage we add exactly the missing edges required by the outgoing and incoming conditions, while keeping uniqueness. That means we construct an increasing sequence of graphs inductively: \[(F_0,\phi_0)\subseteq (F_1,\phi_1)\subseteq (F_2,\phi_2)\subseteq\cdots .\]

Assume that \((F_n,\phi_n)\) has been constructed. For \(z\in F_n^0\) and \(j\in\mathbb{N}\), define \[O_n(z,j)= \{h\in F_n^1\mid s_{F_n}(h)=z,\; \operatorname{tag}(\phi_n^1(h))=j\}.\] Thus \(O_n(z,j)\) records the edges in \(F_n\) which start at \(z\) and have tag \(j\). Similarly, for \(z\in F_n^0\) and \(\mathtt{f}\in E^{\mathrm{st}}\) with \(r(\mathtt{f})=\phi_n^0(z)\), define \[I_n(z,\mathtt{f})= \{h\in F_n^1\mid r_{F_n}(h)=z,\; \operatorname{st}(\phi_n^1(h))=\mathtt{f}\}.\] Thus \(I_n(z,\mathtt{f})\) records the edges in \(F_n\) which end at \(z\) and have structure edge \(\mathtt{f}\). We shall make sure that, for every \(n\), \[\label{eq:In-On} |O_n(z,j)|\leq 1\qquad \text{and}\qquad |I_n(z,\mathtt{f})|\leq 1.\tag{4}\] These inequalities hold for \(n=0\).

We now identify the defects at stage \(n\). The outgoing defects are \[D_n^+=\{(z,j)\mid z\in F_n^0,\;1\leq j\leq \omega(\phi_n^0(z)), \;O_n(z,j)=\emptyset\}.\] The incoming defects are \[D_n^-=\{(z,\mathtt{f})\mid z\in F_n^0,\;\mathtt{f}\in E^{\mathrm{st}}, \;r(\mathtt{f})=\phi_n^0(z),\;I_n(z,\mathtt{f})=\emptyset\}.\] So \(D_n^+\) lists the missing outgoing tags, while \(D_n^-\) lists the missing incoming structure edges.

We first repair the outgoing defects. Let \((z,j)\in D_n^+\). Since \(1\leq j\leq \omega(\phi_n^0(z))\), there exists a structured edge \(\mathtt{f}(z,j)\in E^{\mathrm{st}}\) such that \[s(\mathtt{f}(z,j))=\phi_n^0(z),\quad\text{where}\quad \omega(\mathtt{f}(z,j))\geq j.\] Add a new vertex \(z_j^+\) and a new edge \(f_{z,j}^+\) from \(z\) to \(z_j^+\). Define \[\phi_{n+1}^0(z_j^+)=r(\mathtt{f}(z,j))\quad\text{and}\quad \phi_{n+1}^1(f_{z,j}^+)=\mathtt{f}(z,j)_j.\]

We next repair the incoming defects. Let \((z,\mathtt{f})\in D_n^-\). Add a new vertex \(z_{\mathtt{f}}^-\) and a new edge \(f_{z,\mathtt{f}}^-\) from \(z_{\mathtt{f}}^-\) to \(z\). Define \[\phi_{n+1}^0(z_{\mathtt{f}}^-)=s(\mathtt{f})\quad\text{and}\quad \phi_{n+1}^1(f_{z,\mathtt{f}}^-)=\mathtt{f}_1.\] All new vertices and new edges are taken to be distinct. On the old graph \(F_n\), set \(\phi_{n+1}=\phi_n\).

We now check that \(\phi_{n+1}:F_{n+1}\to\widehat E\) is a graph homomorphism. For an edge \(f_{z,j}^+\) added to repair an outgoing defect, we have \[\widehat s(\mathtt{f}(z,j)_j)=s(\mathtt{f}(z,j))=\phi_n^0(z)\] and \[\widehat r(\mathtt{f}(z,j)_j)=r(\mathtt{f}(z,j))=\phi_{n+1}^0(z_j^+).\] For an edge \(f_{z,\mathtt{f}}^-\) added to repair an incoming defect, we have \[\widehat s(\mathtt{f}_1)=s(\mathtt{f})=\phi_{n+1}^0(z_{\mathtt{f}}^-)\] and \[\widehat r(\mathtt{f}_1)=r(\mathtt{f})=\phi_n^0(z).\] Thus the source and range are respected for every new edge, and hence \(\phi_{n+1}\) is a graph homomorphism.

We now verify that the inequalities 4 are preserved. Let \(z\in F_n^0\) be an old vertex. For each missing outgoing tag \(j\), we add exactly one edge starting at \(z\) with tag \(j\), and we add no edge for a tag already represented in \(O_n(z,j)\). Hence \(|O_{n+1}(z,j)|\leq 1\). Similarly, for each missing incoming structure edge \(\mathtt{f}\), we add exactly one edge ending at \(z\) with structure edge \(\mathtt{f}\), and we add no edge for a structure edge already represented in \(I_n(z,\mathtt{f})\). Hence \(|I_{n+1}(z,\mathtt{f})|\leq 1\).

It remains to check the new vertices. First suppose that \(z'=z_j^+\). Then \(\phi_{n+1}^0(z')=r(\mathtt{f}(z,j))\), and the only edge of \(F_{n+1}\) ending at \(z'\) is \(f_{z,j}^+\). Hence, for any \(\mathtt{g}\in E^{\mathrm{st}}\) with \(r(\mathtt{g})=\phi_{n+1}^0(z')\), \[I_{n+1}(z',\mathtt{g})= \begin{cases} \{f_{z,j}^+\},&\text{if }\mathtt{g}=\mathtt{f}(z,j),\\ \emptyset,&\text{if }\mathtt{g}\neq \mathtt{f}(z,j). \end{cases}\] No edge of \(F_{n+1}\) starts at \(z'\), so \[O_{n+1}(z',\ell)=\emptyset\] for \(1\leq \ell\leq \omega(\phi_{n+1}^0(z'))\). Thus the inequalities 4 hold at \(z'\).

Now suppose that \(z'=z_{\mathtt{f}}^-\). Then \(\phi_{n+1}^0(z')=s(\mathtt{f})\), and the only edge of \(F_{n+1}\) starting at \(z'\) is \(f_{z,\mathtt{f}}^-\). Its image is \(\mathtt{f}_1\). Hence, for \(1\leq \ell\leq \omega(\phi_{n+1}^0(z'))\), \[O_{n+1}(z',\ell)= \begin{cases} \{f_{z,\mathtt{f}}^-\},&\text{if }\ell=1,\\ \emptyset,&\text{if }\ell\neq 1. \end{cases}\] No edge of \(F_{n+1}\) ends at \(z'\), so \(I_{n+1}(z',\mathtt{g})=\emptyset\) for every \(\mathtt{g}\in E^{\mathrm{st}}\) with \(r(\mathtt{g})=\phi_{n+1}^0(z')\). Thus the inequalities 4 also hold at \(z'\).

This completes the induction. Put \[F=\bigcup_{n\geq 0}F_n\qquad\text{and}\qquad \phi=\bigcup_{n\geq 0}\phi_n.\] Since every new vertex is attached by an edge to a vertex already present, the graph \(F\) is connected. The original edge \(e_i\) from \(u\) to \(v\) remains in \(F\), and \(\phi^1(e_i)=\mathtt{e}_i\).

It remains only to check that \((F,\phi)\) is a representation graph. Let \(z\in F^0\), and choose \(n\) such that \(z\in F_n^0\). If \(1\leq j\leq \omega(\phi^0(z))\), then either \(O_n(z,j)\neq\emptyset\), or \((z,j)\in D_n^+\). In the second case, the edge \(f_{z,j}^+\) is added in \(F_{n+1}\). Thus there is at least one edge \(h\in s_F^{-1}(z)\) with \(\operatorname{tag}(\phi^1(h))=j\). There is at most one such edge because \(|O_m(z,j)|\leq 1\) for every finite \(m\), and any two edges of \(F\) already occur together in some \(F_m\).

Similarly, let \(\mathtt{f}\in E^{\mathrm{st}}\) satisfy \(r(\mathtt{f})=\phi^0(z)\). Then either \(I_n(z,\mathtt{f})\neq\emptyset\), or \((z,\mathtt{f})\in D_n^-\). In the second case, the edge \(f_{z,\mathtt{f}}^-\) is added in \(F_{n+1}\). Thus there is at least one edge \(h\in r_F^{-1}(z)\) with \(\operatorname{st}(\phi^1(h))=\mathtt{f}\). There is at most one such edge by the same finite-stage argument.

Therefore \((F,\phi)\) satisfies both the outgoing and incoming conditions in Definition 2. Hence \((F,\phi)\) is a connected representation graph of \((E,\omega)\) containing the prescribed lift \(e_i\) of \(\mathtt{e}_i\). ◻

Proof. Put \(A=L_K(E,\omega)\). Choose \(1\leq i\leq\omega(\mathtt{f})\), and put \(\mathtt{v}_0=s(\mathtt{f})\), \(\mathtt{v}_1=r(\mathtt{f})\) and \(a=\mathtt{f}_i\). Thus \(\mathtt{v}_0\neq \mathtt{v}_1\), and we get \(a^2=(a^*)^2=0\). It follows that \[U=1+2a\quad\text{and}\quad W=1+2a^*\] are in \(A^\times\), with \(U^{-1}=1-2a\) and \(W^{-1}=1-2a^*\) respectively.

By Lemma 1, there is a connected representation graph \((F,\phi)\) of \(E\) and an edge \(f_i\) from \(v_0\) to \(v_1\) in \(F\) such that \(\phi^1(f_i)=\mathtt{f}_i\). Then \(\phi^0(v_0)=\mathtt{v}_0\) and \(\phi^0(v_1)=\mathtt{v}_1\).

Let \(V_F\) be the right \(A\)-module associated to \((F,\phi)\), with basis \(F^0\). Since \(s(f_i)=v_0\) and \(r(f_i)=v_1\) and \(\phi^1(f_i)=\mathtt{f}_i\), it is straightforward to check that \[v_0\cdot a=v_1,\qquad v_1\cdot a^*=v_0, \qquad v_1\cdot a=0, \quad\text{ and }\quad v_0\cdot a^*=0.\]

Set \(\mathcal{W}=Kv_0\oplus Kv_1\), and for \(\xi=\lambda v_0+\mu v_1\in\mathcal{W}\), write \([\xi]_{(v_0,v_1)}=(\lambda,\mu)\) for its row coordinate vector with respect to the ordered basis \((v_0,v_1)\). For \(r\in A\) such that \(\mathcal{W} \cdot r \subseteq \mathcal{W}\), let \(M_r\in \mathrm{M}_2(K)\) be the matrix of the right action of \(r\) on \(\mathcal{W}\), determined by \[[\xi\cdot r]_{(v_0,v_1)}=[\xi]_{(v_0,v_1)}M_r,\quad\text{where}\quad\xi\in\mathcal{W}.\] Since \(v_0\cdot U=v_0+2v_1\) and \(v_1\cdot U=v_1\), we have \[(1,0)M_U=(1,2)\qquad\text{and}\qquad (0,1)M_U=(0,1).\] Hence \[M_U= \begin{pmatrix} 1&2\\ 0&1 \end{pmatrix}.\] Similarly, \[M_W= \begin{pmatrix} 1&0\\ 2&1 \end{pmatrix}.\] Thus, with respect to row coordinates on the ordered basis \((v_0,v_1)\), the restrictions of the right actions of \(U\) and \(W\) to \(\mathcal{W}\) are presented by the following matrices: \[M_U= \begin{pmatrix} 1&2\\ 0&1 \end{pmatrix} \quad\text{and}\quad M_W= \begin{pmatrix} 1&0\\ 2&1 \end{pmatrix}.\]

Since \(\operatorname{char}K=0\), Sanov’s theorem [11] applies to these matrices: the subgroup \(\langle M_U,M_W\rangle\leq \operatorname{GL}_2(K)\) is free of rank two. Let \(F_2=\langle X,Y\rangle\) be the free group of rank two on \(X\) and \(Y\). Since \(U,W\in A^\times\), the universal property of \(F_2\) gives a unique group homomorphism \(\theta:F_2\longrightarrow A^\times\) such that \(\theta(X)=U\) and \(\theta(Y)=W\). Thus, if \(w(X,Y)\in F_2\) is a reduced word in \(X\) and \(Y\), then \(\theta(w(X,Y))=w(U,W)\). In particular, we have \(\theta(1_{F_2})=1\).

Let \(w(X,Y)\in F_2\) be nontrivial reduced word in \(X\) and \(Y\). As \(M_U\) and \(M_W\) are the generators of a free group of rank 2, we get that \(w(M_U,M_W)\neq I_2\). On the other hand, the action of \(w(U,W)\) on \(\mathcal{W}\), with respect to row coordinates in the ordered basis \((v_0,v_1)\), is given by the matrix \(w(M_U,M_W)\). Hence \(w(U,W)\neq 1\), since \(1\) acts on \(\mathcal{W}\) as \(I_2\). Therefore no nontrivial reduced word lies in \(\ker\theta\), and so \(\ker\theta=\{1\}\). Hence \(\langle U,W\rangle\) is a free group of rank two in \(A^\times\). ◻

Proof. The equivalence \((2)\Leftrightarrow(3)\) is exactly [1]. Thus \(A\) is a domain precisely in the three cases listed in \((3)\).

We prove \((3)\Rightarrow(4)\). In case \((3)(a)\), \(A=K\mathtt{v}\), hence \(A^\times=K^\times\mathtt{v}\). In case \((3)(b)\), relations (W3) and (W4) implies that \(\mathtt{e}_1\mathtt{e}_1^*=\mathtt{v}=\mathtt{e}_1^*\mathtt{e}_1\). Hence the mapping \(x\mapsto \mathtt{e}_1\) induces an isomorphism \(K[x,x^{-1}]\cong A\), and \(A^\times=K^\times\mathtt{e}_1^{\mathbb{Z}}\). In case \((3)(c)\), by Theorem 1, \(A^\times=K^\times\mathtt{v}\). Hence \((3)\Rightarrow(4)\). Since the groups in \((4)\) are abelian, \((4)\Rightarrow(1)\).

It remains to prove \((1)\Rightarrow(3)\). Assume by contradiction that \((E,\omega)\) is not one of the three graphs in \((3)\). First suppose that \(|E^0|>1\). Since \(E\) is connected, there is \(\mathtt{f}\in E^{\mathrm{st}}\) with \(s(\mathtt{f})\neq r(\mathtt{f})\). Choose \(1\leq i\leq\omega(\mathtt{f})\), and put \[a=\mathtt{f}_i,\qquad \mathtt{v}_0=s(\mathtt{f}),\qquad\text{and}\qquad \mathtt{v}_1=r(\mathtt{f}).\] Then \(\mathtt{v}_0\neq\mathtt{v}_1\), and so \(a^2=0=(a^*)^2\). Hence \[U=1+a\quad\text{and}\quad W=1+a^*\] are invertible in \(A\), with inverses \(1-a\) and \(1-a^*\).

By Lemma 1, there is a representation graph \((F,\phi)\) containing an edge \(f_i\) from \(v_0\) to \(v_1\) such that \(\phi^1(f_i)=\mathtt{f}_i\). Let \(V_F\) be the right \(A\)-module associated to \((F,\phi)\). Since \[\widehat s(\mathtt{f}_i)=\mathtt{v}_0\quad\text{and}\quad \widehat r(\mathtt{f}_i)=\mathtt{v}_1,\] the action on \(V_F\) implies that \[v_0\cdot a=v_1,\qquad v_1\cdot a^*=v_0,\qquad v_1\cdot a=0,\qquad v_0\cdot a^*=0.\] Consequently \[v_0\cdot aa^*=(v_0\cdot a)\cdot a^*=v_1\cdot a^*=v_0 \qquad \text{whereas} \qquad v_0\cdot a^*a=(v_0\cdot a^*)\cdot a=0.\] Thus \(aa^*\neq a^*a\) and so \(UW\neq WU\), which implies that \(A^\times\) is not abelian.

Now suppose that \(|E^0|=1\). Since \((E,\omega)\) is not one of the graphs in \((3)\), it is neither an isolated vertex, nor \(R_1\), nor an LV-rose. By Corollary 1, \(A^\times\) contains a non-cyclic free subgroup. In particular, \(A^\times\) is not abelian. This proves that \((1)\Rightarrow(3)\), and hence \((1)\)–\((4)\) are equivalent.

If the equivalent conditions \((1)\)–\((4)\) hold, then \(A^\times\) is abelian, so it contains no non-cyclic free subgroup. Conversely, suppose that these conditions fail. If \(|E^0|>1\), then there is \(\mathtt{f}\in E^{\mathrm{st}}\) with \(s(\mathtt{f})\neq r(\mathtt{f})\), and Proposition 2 shows that there is a free subgroup of rank two in \(A^\times\). If \(|E^0|=1\), then Corollary 1 yields a free subgroup of rank two inside \(A^\times\). This proves the equivalence of \((1)\)–\((5)\). ◻