Proof. To prove the contrapositive of item (1), suppose that $H=\mathrm {Stab}_G(y)$ is infinite for some vertex y of X. Let $B_n$ be the ball of radius $n \in \mathbb {N}$ about y. Since X is locally finite, the following inductive argument shows that there are infinitely many distinct elements of H that fix each $B_n$ pointwise.

By the inductive hypothesis, we may assume that H has infinitely many elements that fix $B_n$ . Infinitely many of these elements, say $\{g_i : i\in I\}$ , induce the same permutation on the finitely many vertices in $B_{n+1}\setminus B_n$ . The set $\{g_ig_j^{-1} : i,j\in I\}$ then contains infinitely many distinct elements of H, each of which fixes $B_{n+1}$ pointwise. Hence, we may choose a sequence $\{g_n\}$ of distinct nontrivial elements of H such that $g_n$ fixes $B_n$ pointwise. Thus, $\{g_n\}$ converges to 1, so G is not discrete.

To prove item (2), suppose that $\mathrm {Stab}_G(y)$ is finite for some vertex y and suppose that $\{g_i\}$ is a sequence of elements of G converging to the identity. The sequence of vertices $\{g_i \cdot y \}$ converges to y, so there exists an N such that $g_i \in \mathrm {Stab}_G(y)$ for each $i\geq N$ . It follows that the sequence $\{g_i\}$ is eventually $1$ , and hence G is discrete.

Proof. We first note that there is an ${\mathrm {SL}_2}(K)$ -equivariant isomorphism between the boundary $\partial T_K$ and the projective line $\mathbb {P}^1(K)$ [Reference Serre26]. Hence, after conjugation if necessary, we may assume that A fixes the end of $T_K$ corresponding to the eigenvector $(1,0)$ or, equivalently, A is upper triangular. A standard trace computation (similar to the proof of [Reference Beardon2, Theorem 4.3.5(i)]) then shows that $\mathrm {tr}([A,B])=2$ if and only if either B is also upper triangular or A is diagonal and B is lower triangular. In either case, A and B fix a common end of $T_K$ .

Proof. Let $q=p^r$ be the size of the residue field of K. If $X\in {\mathrm {SL}_2}(K)$ has finite order n coprime to p, then [Reference Conder and Schillewaert9, Proposition 3.3] shows that $n \mid q \pm 1$ . If X has order $p^k$ , then X is unipotent and hence $k=1$ when $\mathrm {char}(K)>0$ by [Reference Lubotzky20, page 964], and ${(p-1)p^{k-1}\le 2[K:\mathbb {Q}_p]}$ when $\mathrm {char}(K)=0$ by [Reference Serre25, Proposition 17, page 78] since either K or a quadratic extension of K must contain a $p^k$ th root of unity. Otherwise, if X has order $p^kn$ where $p\nmid n$ , then $X^{p^k}$ and $X^n$ have orders n and $p^k$ , respectively, so the result follows from the two cases above.

Proof of Theorem A.

Suppose that $G=\langle A,B \rangle \le {\mathrm {SL}_2}(K)$ is discrete and violates (1-1), that is,

$$ \begin{align*}{\operatorname{min}}\{v(\mathrm{tr}^2(A)-4), v(\mathrm{tr}([A,B])-2) \}> M_K\ge 0.\end{align*} $$

If A (respectively $[A,B]$ ) is hyperbolic, then [Reference Cassels8, Ch. 2, Lemma 1.4] and Pro- position 2.4 imply that $v(\mathrm {tr}^2(A)-4)=v(\mathrm {tr}^2(A))<0$ (respectively $v(\mathrm {tr}([A,B])-2)= v(\mathrm {tr}([A,B]))<0$ ), which is a contradiction. Since G contains no infinite order elliptic elements by Lemma 2.1, we deduce that A and $[A,B]$ are both elliptic of finite order.

Now, after applying the well-known trace equality $\mathrm {tr}^2(A)=\mathrm {tr}(A^2)+2$ , it follows from the definition of $M_K$ that $\mathrm {tr}(A)=\pm 2$ and $\mathrm {tr}([A,B])=2$ . Hence, A has characteristic polynomial $x^2 \pm 2x +1=(x \pm 1)^2$ and repeated eigenvalue $\pm 1$ , so A fixes exactly one end $\eta $ of $T_K$ . Lemma 2.2 then shows that B fixes $\eta $ , thus G fixes an end of $T_K$ .

Proof of Theorem B.

Let $G=\langle A, B \rangle $ be a discrete nonelementary subgroup of ${\mathrm {SL}_2}(K)$ , where either $K=\mathbb {Q}_p$ or G contains no elements of order p. Since $v(2)$ and $v(4)$ are nonnegative, Proposition 2.4 and the ultrametric inequality show that (1-2) holds with strict inequality if A or $[A,B]$ is hyperbolic. Hence, we may assume that A and $[A,B]$ are elliptic.

If B is also elliptic, then the fixed point sets of A and B are disjoint since G is nonelementary. The proof of [Reference Culler and Morgan10, 1.5] shows that the axes of $AB$ and $A^{-1}B^{-1}$ translate in the same direction along the unique geodesic between the fixed point sets of A and B. It follows from [Reference Culler and Morgan10, 1.8] that $[A,B]$ is hyperbolic, which is a contradiction. Hence, B is hyperbolic.

Note that A and $BA^{-1}B^{-1}$ must fix a common vertex, as otherwise $[A,B]$ is hyperbolic by [Reference Culler and Morgan10, 1.5]. Since A cannot fix an end of the translation axis $\mathrm {Ax}(B)$ of B, this implies that A fixes a subpath of $\mathrm {Ax}(B)$ of finite length $\Delta \ge l(B)$ .

Since G is discrete, A has finite order and $\Delta =l(B)$ by Lemma 2.1 and [Reference Conder and Schillewaert9, Lemma 3.10]. Proposition 2.4 implies that $\Delta \ge 2$ and hence the fixed point set of A cannot be a vertex or an edge. Thus, A fixes two ends of $T_K$ by [Reference Conder and Schillewaert9, Proposition 3.4] and is hence diagonalisable over K. Let $\lambda , \lambda ^{-1} \in K$ be the eigenvalues of A which, since G is nonelementary, must be distinct n th roots of unity for some $n>2$ . By Hensel’s lemma [Reference Serre25, II Section 4, Proposition 7], each n th root of unity in K is the unique lift of an n th root of unity modulo $\pi $ , where $\pi $ is the uniformiser of K. It follows that $\lambda \neq \lambda ^{-1} (\mod\pi) $ and hence ${v(\mathrm {tr}^2(A)-4)}=2v(\lambda -\lambda ^{-1})=0$ , so we obtain equality in (1-2).

Proof. We start by showing that G must contain a hyperbolic element g. Suppose for a contradiction that all elements of G are elliptic. Note that $\mathrm {Fix}(g_i)\cap \mathrm {Fix}(g_j) \neq \varnothing $ for each pair of distinct elements $g_i,g_j \in G$ , as otherwise, [Reference Culler and Morgan10, 1.5] shows that $g_ig_j$ is hyperbolic. Hence, G fixes either a vertex or an end by [Reference Tits, Bass, Cassidy and Kovacic28, Lemma 1.6], which is a contradiction.

Let $\eta ^+,\eta ^-$ be the two ends of $\mathrm {Ax}(g)$ . We may suppose that the axis of every hyperbolic element of G has an end in common with $\mathrm {Ax}(g)$ , as otherwise, there is nothing to prove. Since G is nonelementary, there are hyperbolic elements $h_1, h_2$ such that $h_1$ fixes $\eta ^+$ (but not $\eta ^-$ ) and $h_2$ fixes $\eta ^-$ (but not $\eta ^+$ ). If $\mathrm {Ax}(h_1)$ and $\mathrm {Ax}(h_2)$ have pairwise distinct ends, then the proof is complete, so we may suppose that $\mathrm {Ax}(h_1)$ and $\mathrm {Ax}(h_2)$ have a common end $\zeta \notin \{\eta ^+, \eta ^-\}$ . Since the ends $\eta ^+$ and $\zeta $ of $\mathrm {Ax}(h_1)$ are distinct from the ends $h_2\cdot \eta ^+$ and $\eta ^-$ of $\mathrm {Ax}(h_2gh_2^{-1})=h_2\cdot \mathrm {Ax}(g)$ , this proves the lemma.

Proof. Let $\eta ^-$ and $\eta ^+$ , respectively, be the repelling and attracting ends of $\mathrm {Ax}(h)$ . Without loss of generality, we may suppose that g fixes $\eta ^-$ .

If g is elliptic, then g fixes a halfray R on $\mathrm {Ax}(h)$ . If x is a point on R, then $h^igh^{-i}$ fixes x for each $i\geq 0$ . Since $\mathrm {Stab}_G(x)$ is finite by Lemma 2.1, we obtain that $h^igh^{-i}=h^jgh^{-j}$ for some distinct i and j. Thus, g commutes with $h^k$ , where $k=i-j$ .

If g is hyperbolic, then $\mathrm {Ax}(g)$ intersects $\mathrm {Ax}(h)$ in at least a halfray R and we may assume (by inverting g if necessary) that $\eta ^-$ is also the repelling end of $\mathrm {Ax}(g)$ . Let $x \in R$ and observe that $gh^ig^{-1}h^{-i}$ fixes x for each $i\geq 0$ . A similar argument to above then shows that g commutes with $h^k$ for some integer k.

In either case, $g \cdot \eta ^+ = gh^k \cdot \eta ^+ = h^k (g \cdot \eta ^+)$ , so $g \cdot \eta ^+ \in \{\eta ^-,\eta ^+\}$ . Hence, g fixes $\eta ^+$ .

Proof. By Lemma 2.1, h must be hyperbolic. Let $\eta ^-$ and $\eta ^+$ be the ends of $\mathrm {Ax}(h)$ and suppose for a contradiction that $\langle g_i,h \rangle $ is elementary for every i. By Lemma 3.2, $g_i$ cannot fix precisely one of $\eta ^-$ or $\eta ^+$ , and hence $g_i$ stabilises the set $\{\eta ^-,\eta ^+\}$ . Thus, G stabilises $\{\eta ^-,\eta ^+\}$ and is hence elementary, which is a contradiction.

Proof. Since G is a discrete subgroup of a Hausdorff group, it is closed. Hence, by [Reference Caprace and De Medts7, Remark 2.3 and Lemma 2.4], there exists a subtree of T on which the action of G is cocompact. Since G acts properly on $T_K$ , it follows from [Reference Bridson and Haefliger5, Corollary I.8.11] that G is finitely presented.

Proof. Let G be a discrete elementary subgroup of ${\mathrm {SL}_2}(K)$ . If G fixes a vertex of $T_K$ , then G is finite by Lemma 2.1. So suppose that G fixes precisely one end of $T_K$ . By Lemma 3.2, G consists entirely of finite order elliptic elements. Using the identification of $\partial T_K$ with the projective line $\mathbb {P}^1(K)$ [Reference Serre26, page 72], we may conjugate G into a group of upper triangular matrices. Its unipotent elements form an abelian normal subgroup H of G, such that $G/H$ can be identified with a finite subgroup of the collection of roots of unity of K; for instance, see the argument given in [Reference Beardon2, page 87]. (Note that, in this case, H is trivial if $\mathrm {char}(K)=0$ , and is an elementary abelian p-group if $\mathrm {char}(K)=p>0$ .) Hence, G is virtually abelian. Finally, we suppose that G stabilises a pair of ends of $T_K$ . Thus, G contains a subgroup H of index at most two which pointwise fixes these two ends. Again using the identification of $\partial T_K$ with $\mathbb {P}^1(K)$ , we may assume by conjugation that H consists entirely of diagonal matrices and is therefore abelian.

Conversely, suppose that G is discrete and virtually abelian. If G is finite, then it fixes a vertex of $T_K$ by [Reference Tits, Bass, Cassidy and Kovacic28, 2.3.1], so suppose that G is infinite. In particular, G contains an infinite abelian normal subgroup H. If H contains a hyperbolic element h, then Lemmas 2.2 and 3.2 show that every element of H fixes both ends $\eta ^\pm $ of $\mathrm {Ax}(h)$ . Moreover, if $g\in G\setminus H$ , then since $H\trianglelefteq G$ , we obtain that $(g^{-1}Hg)(\eta ^{\pm }) = \eta ^{\pm }$ , whence g stabilises $\{\eta ^+, \eta ^-\}$ . However, if H contains only elliptic elements, then [Reference Tits, Bass, Cassidy and Kovacic28, Lemma 1.6] implies that H fixes either a vertex or an end of $T_K$ . The former option contradicts Lemma 2.1, whence H fixes an end of $T_K$ and a similar argument to the above then shows that G fixes this same end.

Proof. If g is elliptic, then let $x \in \mathrm {Fix}(g)$ . Using the topology of pointwise convergence, we may choose a positive integer N such that $d(g_n\cdot x, x)=d(g_n\cdot x, g \cdot x)<l_{\operatorname {min}}$ for each $n \ge N$ . It follows that $g_n$ fixes x for each $n \ge N$ .

However, if g is hyperbolic, then assume for a contradiction that there is a subsequence $\{h_n\}$ of $\{g_n\}$ that consists only of elliptic elements and converges to g. Let x be a point of T and let m be the midpoint of $[x, g\cdot x]$ . Using the topology of pointwise convergence, we may choose a positive integer N such that $d(h_n\cdot x,g \cdot x)<\tfrac 12l(g)$ and $d(h_n \cdot m, g \cdot m)<\tfrac 12l(g)$ for each $n \ge N$ . By [Reference Culler and Morgan10, 1.3(iii)], the midpoint $m_n$ of $[x, h_n\cdot x]$ is fixed by $h_n$ for every n. Note that $d(m_n,m)\leq \tfrac 12 d(h_n \cdot x, g\cdot x)<\tfrac 14l(g)$ by [Reference Bridson and Haefliger5, Proposition II.2.2] and thus

$$ \begin{align*}d(h_n\cdot m,m)\leq d(h_n\cdot m,h_n\cdot m_n)+d(h_n \cdot m_n,m_n)+d(m_n,m)<\tfrac{1}{2}l(g)\end{align*} $$

for each $n \ge N$ . Hence, $d(g\cdot m,m)\leq d(g\cdot m,h_n\cdot m)+d(h_n\cdot m,m)<l(g)$ for sufficiently large n, which is a contradiction.

Proof. By Lemma 4.1(1), the sequence $\{g_n\}$ is eventually elliptic. It follows from Lemma 2.1 that $\{g_n\}$ eventually consists of finite order elements. Since the trace function is continuous, Corollary 2.6 shows that $\{\mathrm {tr}(g_n)\}$ is eventually constant.

Proof. Suppose for a contradiction that G is nonelementary. By Lemma 3.1, there are hyperbolic elements $h_1, h_2 \in G$ such that the ends of $\mathrm {Ax}(h_1)$ and $\mathrm {Ax}(h_2)$ are distinct. Without loss of generality, we may assume that $h_1$ and $h_2$ translate in the same direction along the (possibly empty) finite path $\mathrm {Ax}(h_1) \cap \mathrm {Ax}(h_2)$ . There is some positive integer k such that $\mathrm {Ax}(h_1)$ and $\mathrm {Ax}(h_2^kh_1^{-1}h_2^{-k})=h_2^k\cdot \mathrm {Ax}(h_1)$ intersect along a (possibly empty) path of length strictly less than $l(h_1)$ , and hence [Reference Culler and Morgan10, 1.5 and 3.4] show that $[h_1,h_2^k]$ is hyperbolic. By Lemma 4.1(2), there is a sufficiently large positive integer N such that the corresponding elements $h_{1_N}$ , $h_{2_N}$ and $[h_{1_N},h_{2_N}^k]$ of $G_N$ are hyperbolic. Since $[h_{1_N},h_{2_N}^k]$ is hyperbolic, $\mathrm {Ax}(h_{1_N})$ and $\mathrm {Ax}(h_{2_N})$ must have finite (or empty) overlap by [Reference Culler and Morgan10, Corollary 2.3], which contradicts the fact that $G_N$ is elementary.

Proof. Suppose first that $g_n$ and $h_n$ fix a common end of $T_K$ for all sufficiently large n. By Lemma 2.2, $\mathrm {tr}([g_n,h_n])=2$ for all sufficiently large n, and so $\mathrm {tr}([g,h])=2$ since the trace function is continuous. Hence, Lemma 2.2 shows that g fixes an end of $\mathrm {Ax}(h)$ .

Conversely, suppose that g fixes an end of $\mathrm {Ax}(h)$ . By Lemma 2.2, $\mathrm {tr}([g,h])=2$ and hence $[g,h]$ is elliptic by Proposition 2.4. Corollary 4.3 then shows that $\mathrm {tr}([g_n,h_n])=2$ for all sufficiently large n. Since $h_n$ is hyperbolic for sufficiently large n by Lemma 4.1 $(2)$ , another application of Lemma 2.2 proves the result.

Proof. Since $\langle g,h_1 \rangle $ and $\langle g,h_2 \rangle $ are elementary, g stabilises the ends of both $\mathrm {Ax}(h_1)$ and $\mathrm {Ax}(h_2)$ . Hence, $g^2$ fixes four ends of $T_K$ and, since $\partial T_K$ can be identified with the projective line $\mathbb {P}^1(K)$ [Reference Serre26, page 72], it follows that $g^2=1$ .

Proof. Let $\{g_i\}$ be a sequence of elements of G converging to $1$ . By Lemma 3.1, we may choose hyperbolic elements $h_1, h_2 \in G$ whose axes have no end in common. Observe that the sequences $\{[g_i,h_1]\}$ and $\{[g_i,h_2]\}$ both converge to $1$ . For sufficiently large i and n, it follows that ${\operatorname {min}}\{v(\mathrm {tr}^2(g_{i_n})-4), v(\mathrm {tr}([g_{i_n},h_{j_n}])-2)\}>M_K$ for $j \in \{1,2\}$ . Theorem A hence implies that the subgroups $\langle g_{i_n}, h_{1_n} \rangle $ and $\langle g_{i_n}, h_{2_n}\rangle $ of $G_n$ are elementary for sufficiently large i and n. By Lemma 4.4, the subgroups $\langle g_i, h_1 \rangle $ and $\langle g_i, h_2 \rangle $ of G are also elementary for sufficiently large i. Hence, $\{g_i\}$ is eventually constant by Lemma 4.6, so G is discrete.

Proof of Proposition 1.2.

Let G be a nonelementary subgroup of ${\mathrm {SL}_2}(K)$ . If G is discrete, then so is every subgroup of G. So suppose that every two-generator subgroup of G is discrete, and that $\{g_i\}$ is a sequence of elements of G converging to $1$ . By Lemma 3.1, we may choose hyperbolic elements $h_1, h_2 \in G$ whose axes have pairwise distinct ends. For sufficiently large i, the following inequality holds for each $j \in \{1,2\}$ :

$$ \begin{align*} {\operatorname{min}}\{v(\mathrm{tr}^2(g_i)-4), v(\mathrm{tr}([g_i,h_j]-2)\}>M_K.\end{align*} $$

Hence, Theorem A shows that $\langle g_i, h_1\rangle $ and $\langle g_i, h_2\rangle $ are both elementary for sufficiently large i. By Lemma 4.6, $\{g_i\}$ is eventually constant and hence G is discrete.

Proof of Proposition 1.3.

If G is discrete, then every cyclic subgroup of G is discrete. Hence, we may suppose that every cyclic subgroup of G is discrete. Since G acts by isometries on a locally finite simplicial building X of type $\tilde {A}_{n-1}$ [Reference Bruhat and Tits6, 2.2.8 and 7.4.11], Lemma 2.1 shows that every element in G that fixes a vertex of X has finite order. Thus, every vertex stabiliser in G is periodic, that is, it consists only of finite order elements. Moreover, every vertex stabiliser in G has finite exponent by Lemma 2.5. Thus, every vertex stabiliser in G is finite by [Reference Wehrfritz29, Theorem 9.1(ii) and (iii)], so G is discrete by Lemma 2.1.

Proof of Theorem C.

We first prove that $\phi $ is an isomorphism. Since $\phi $ is surjective by construction, it suffices to show that $\phi $ is injective.

Let $A \in \Gamma $ be nontrivial. If A is hyperbolic, then the proof of Lemma 3.1 shows that we may choose a hyperbolic element $B \in \Gamma $ such that $\mathrm {Ax}(A)$ and $\mathrm {Ax}(B)$ have pairwise distinct ends. If A is elliptic, then there exists a hyperbolic element $B \in \Gamma $ such that A does not stabilise the set of ends of $\mathrm {Ax}(B)$ . In either case, there exists a hyperbolic element $B \in \Gamma $ such that $\langle A, B \rangle $ is nonelementary. By Lemma 3.5, $\langle \phi _n(A), \phi _n(B) \rangle $ is then a nonelementary subgroup of $G_n$ for sufficiently large n.

Now suppose for a contradiction that $\phi (A)=1$ . The sequence $\{\phi _n(A)\}$ converges to $1$ , so for sufficiently large n, we obtain

$$ \begin{align*}{\operatorname{min}}\{v(\mathrm{tr}^2(\phi_n(A))-4),v(\mathrm{tr}([\phi_n(A),\phi_n(B)])-2)\}>M_K.\end{align*} $$

Theorem A then implies that $\langle \phi _n(A), \phi _n(B) \rangle $ is elementary for sufficiently large n, which gives the desired contradiction. Hence, $\phi $ is an isomorphism.

Now suppose that $\{g_i=\phi (\gamma _i)\}$ is a sequence of elements of G converging to $1$ . Since $G_n$ is isomorphic to $\Gamma $ , Lemma 3.5 shows that $G_n$ is nonelementary for sufficiently large n. For each such n, there exist hyperbolic elements $\phi _n(h_1), \phi _n(h_2) \in G_n$ whose axes have pairwise distinct ends by Lemma 3.1. For sufficiently large i and n, observe that

$$ \begin{align*}{\operatorname{min}}\{v(\mathrm{tr}^2(\phi_n(\gamma_i)-4)), v(\mathrm{tr}([\phi_n(\gamma_i),\phi_n(h_j)])-2)\}>M_K\end{align*} $$

for $j \in \{1,2\}$ . Theorem A hence implies that the subgroups $\langle \phi _n(\gamma _i), \phi _n(h_1) \rangle $ and $\langle \phi _n(\gamma _i), \phi _n(h_2)\rangle $ of $G_n$ are elementary for sufficiently large i and n. Lemma 4.6 thus shows that $\phi _n(\gamma _i)=\pm 1$ for sufficiently large i and n. Since $\phi _n$ and $\phi $ are isomorphisms, it follows that $\{g_i=\phi (\gamma _i)\}$ is eventually 1, whence G is discrete. Moreover, Lemma 3.5 shows that G is nonelementary.

Proof of Theorem D.

By Proposition 4.7, if G is nonelementary, it is also discrete, so we start by proving the former.

For each n, one of the elements in the set ${\{g_{i_n} : 1 \leq i \leq n\} \cup \{ g_{i_n}g_{j_n} : 1\leq i<j \leq r\}}$ must be hyperbolic as otherwise, $G_n$ fixes a point [Reference Serre26, I.6.5 Corollary 2 of Proposition 26] and is hence elementary. We denote this hyperbolic element by $h_n$ . By Lemma 3.3, for each n, there exists $i_n \in \{1,\ldots , r\}$ such that the group $H_n = \langle h_n, g_{i_n} \rangle $ is discrete and nonelementary. Since each $G_n$ is finitely generated, for infinitely many n, we must have picked the same indices to define both generators of $H_n$ . Hence, there exists a subsequence of $\{H_n\}$ converging to some two-generator subgroup H of G. Since G is nonelementary when H is, it thus suffices to prove the statement for $r=2$ .

If every element of $\{g_1, g_2, g_1g_2\}$ is elliptic, then Lemma 4.1(1) shows that every element of $\{g_{1_n}, g_{2_n}, g_{1_n}g_{2_n}\}$ is also elliptic for sufficiently large n. By [Reference Serre26, I.6.5 Corollary 2 of Proposition 26], $G_n$ is elementary, which is a contradiction. Hence, we may choose some hyperbolic element $h \in \{g_1, g_2, g_1g_2\}$ .

If $g \in \{g_1, g_2, g_1g_2\} \backslash \{h\}$ is such that $ghg^{-1}$ fixes an end of $\mathrm {Ax}(h)$ , then Lemmas 4.1(2) and 4.5 show that, for n sufficiently large, $h_n$ is hyperbolic and $g_nh_ng_n^{-1}$ fixes an end of $\mathrm {Ax}(h_n)$ . Thus, Lemma 3.2 shows that $G_n$ is elementary, which is a contradiction. Therefore, there is some $g \in \{g_1, g_2, g_1g_2\}\backslash \{h\}$ such that the hyperbolic elements h and $ghg^{-1}$ have no ends in common, so G is nonelementary (and hence discrete).

To prove that the maps $\psi _n:g_i\mapsto g_{i_n}$ extend to surjective homomorphisms, we argue as in [Reference Jørgensen13, Theorem 2]. Note that Lemma 3.4 shows that G is finitely presented. Therefore, a necessary and sufficient condition for $\psi _n$ to be a homomorphism is that the images of relators in G are equal to the identity.

Let $R_1,\ldots ,R_k$ be a basis for the relations for G. It remains to find a sufficiently large N such that for $n\geq N$ , the corresponding elements in $G_n$ are equal to the identity. By Lemma 3.1, we can find two hyperbolic elements $h_1, h_2 \in G$ whose axes have pairwise distinct ends. It follows from Lemmas 4.1(2) and 4.5 that, for sufficiently large n, there exist hyperbolic elements $h_{1_n}, h_{2_n} \in G_n$ whose axes have pairwise distinct ends.

Since the sequence $\{G_n\}$ converges to G, the corresponding elements $R_{s_n}$ converge to $1$ for each $s\in \{1,\ldots ,k\}$ . By Theorem A, the groups $\langle R_{s_n}, h_{t_n} \rangle $ are elementary for each $s\in \{1,\ldots ,k\}$ , $t\in \{1,2\}$ and sufficiently large n. Hence, Lemma 4.6 shows that $R_{s_n}=1$ for each $s\in \{1,\ldots ,k\}$ and sufficiently large n.

Proof of Proposition 1.5.

Let G be dense in ${\mathrm {SL}_2}(K)$ . Since ${\mathrm {SL}_2}(K)$ contains a two-generator dense subgroup, there exist $g,h\in \overline {G}$ that generate a dense subgroup H of ${\mathrm {SL}_2}(K)$ . Let $\{g_n\}$ and $\{h_n\}$ be sequences of elements of G converging to g and h, respectively. If all but finitely many of the subgroups $H_n =\langle g_n, h_n \rangle $ of G are elementary, then H is elementary by Lemma 4.4. This is a contradiction by [Reference Conder and Schillewaert9, Lemma 6.6], so we may assume without loss of generality that the sequence $\{H_n\}$ consists entirely of nonelementary subgroups of G. If all but finitely many of these subgroups are discrete, then H is discrete by Theorem D, which is again a contradiction. Hence, there exists a positive integer N such that $H_N$ is nonelementary and not discrete. It follows from [Reference Conder and Schillewaert9, Lemmas 6.6 and 6.7] that $H_N$ is dense.