Proof. If this is not the case, then we may find a compact open subgroup $U$ such that $U\cap N=1$ and $U$ contains no open normal subgroup of $G$ . Note that this implies in particular that $G$ is not discrete.

Let $K$ be the normal core of $U$ in $G$ and $\unicode[STIX]{x1D70B}:G\rightarrow G/K$ the quotient map. Note that $\unicode[STIX]{x1D70B}(U)$ is a compact open subgroup of $\unicode[STIX]{x1D70B}(G)$ that contains no non-trivial normal subgroup of $\unicode[STIX]{x1D70B}(G)$ . Moreover, the group $\unicode[STIX]{x1D70B}(G)$ admits $\unicode[STIX]{x1D70B}(N)$ as a discrete normal subgroup, and the quotient $\unicode[STIX]{x1D70B}(G)/\unicode[STIX]{x1D70B}(N)\simeq G/KN$ has a basis of identity neighborhoods consisting of compact open normal subgroups, since it is a quotient of $G/N$ . So, in order to obtain our contradiction, we may assume that $K=1$ .

Consider the family ${\mathcal{F}}$ of open normal subgroups of $G$ containing $N$ . By Proposition 2.5 in [Reference Caprace and MonodCM11], the group $\bigcap _{{\mathcal{F}}}M$ intersects $U$ non-trivially. But the assumption on $G/N$ implies that the subgroup $\bigcap _{{\mathcal{F}}}M$ is equal to $N$ . So, we obtain that $U\cap N$ is non-trivial, which is our contradiction.◻

Proof. By [Reference CornulierCor15, Proposition 2.7], every compact normal subgroup $K$ of $H$ is contained in the polycompact radical $W(G)$ . By definition, the polycompact radical is a subgroup of the locally elliptic radical (see [Reference CornulierCor15, Proposition 2.4(6)]). Since the normalizer of $K$ in $G$ contains $H$ , and is thus cocompact in $G$ , it follows that the union $X$ of the conjugacy classes of elements of $K$ is a compact subset of $G$ . Since $X$ is contained in the locally elliptic radical of $G$ , it follows that the closed subgroup generated by $X$ is compact. This subgroup is also normal in $G$ and contains $K$ , hence the statement.◻

Proof. The subgroup $\unicode[STIX]{x1D6E4}$ is cocompact in the compactly generated group $G$ , so $\unicode[STIX]{x1D6E4}$ is finitely generated. Since $\unicode[STIX]{x1D6E4}$ is a discrete normal subgroup of $N_{G}(\unicode[STIX]{x1D6E4})$ , $\unicode[STIX]{x1D6E4}$ lies in the quasi-center of $N_{G}(\unicode[STIX]{x1D6E4})$ , i.e. every element of $\unicode[STIX]{x1D6E4}$ centralizes an open subgroup of $N_{G}(\unicode[STIX]{x1D6E4})$ . But, since $\unicode[STIX]{x1D6E4}$ is finitely generated, it follows that the centralizer $C_{G}(\unicode[STIX]{x1D6E4})$ is open in $N_{G}(\unicode[STIX]{x1D6E4})$ . If $U$ is a compact open subgroup of $N_{G}(\unicode[STIX]{x1D6E4})$ centralized by $\unicode[STIX]{x1D6E4}$ , then $U\unicode[STIX]{x1D6E4}$ is a subgroup that is open and cocompact in $N_{G}(\unicode[STIX]{x1D6E4})$ , since $\unicode[STIX]{x1D6E4}$ is cocompact in $G$ . Therefore, $U\unicode[STIX]{x1D6E4}$ has finite index in $N_{G}(\unicode[STIX]{x1D6E4})$ . Now the compact subgroup $U$ has a cocompact normalizer in $G$ , so it follows from Proposition 2.3 that $U$ is contained in a compact normal subgroup $K$ of $G$ , and $K\unicode[STIX]{x1D6E4}$ indeed contains a finite index subgroup of $N_{G}(\unicode[STIX]{x1D6E4})$ .◻

Proof. Let $K$ be as in the conclusion of Proposition 2.4. Since $G$ has a discrete polycompact radical, the subgroup $K$ must be finite and it follows that $\unicode[STIX]{x1D6E4}$ has finite index in $K\unicode[STIX]{x1D6E4}\cap N_{G}(\unicode[STIX]{x1D6E4})$ . Since $K\unicode[STIX]{x1D6E4}\cap N_{G}(\unicode[STIX]{x1D6E4})$ has finite index in $N_{G}(\unicode[STIX]{x1D6E4})$ , the conclusion follows.◻

Proof. The polycompact radical $W(G)$ is a topologically characteristic subgroup. Assume that it is non-trivial. Thus, it is dense by hypothesis. Since $G$ is compactly generated, the radical $W(G)$ is closed by [Reference CornulierCor15, Theorem 1.2], so $G=W(G)$ . Since the polycompact radical is a subgroup of the locally elliptic radical (see [Reference CornulierCor15, Proposition 2.4(6)]), it follows that $G$ is locally elliptic. Since $G$ is compactly generated, this implies that $G$ is compact, contradicting the hypotheses. Hence, $W(G)=\{1\}$ .

We next observe that the quasi-center is also a topologically characteristic subgroup. Assume that $\text{QZ}(G)$ is non-trivial. Thus, it is dense by hypothesis. It then follows from [Reference Caprace and MonodCM11, Proposition 4.3] that the compact open normal subgroups of $G$ form a basis of identity neighborhoods. In particular, $W(G)$ is dense, contradicting the first part of the proof.◻

Proof. For both (i) and (ii), the ‘only if’ implications follow from [Reference Caprace and MonodCM11, Theorem E]. Conversely, if $G$ is compactly generated and monolithic such that $\text{Mon}(G)$ is cocompact, then, by the definition of monolithicity, every non-trivial closed normal subgroup of $G$ contains $\text{Mon}(G)$ and is thus cocompact in $G$ . Moreover, $G$ cannot be compact, since a compact tdlc group is profinite, hence residually finite, and can thus not be monolithic. Thus, the condition of part (i) is indeed sufficient. The fact that $G$ has trivial quasi-center then follows from Proposition 2.6.

Assume now that $G$ is compactly generated and monolithic such that $\text{Mon}(G)$ is cocompact and topologically simple. Let $H$ be an open subgroup of finite index in $G$ . Then $H$ contains an open normal subgroup of finite index in $G$ ; in particular, $H$ contains $\text{Mon}(G)$ . Given a non-trivial closed normal subgroup $N$ of $H$ , $N\cap \text{Mon}(G)$ is normal in $\text{Mon}(G)$ . Since the latter is topologically simple by hypothesis, we have $N\geqslant \text{Mon}(G)$ or $N\cap \text{Mon}(G)=1$ . In the latter case we deduce that $N\leqslant \text{C}_{G}(\text{Mon}(G))$ . That centralizer is thus a non-trivial closed normal subgroup of $G$ . It must contain $\text{Mon}(G)$ by definition. This implies that $\text{Mon}(G)$ is self-centralizing and hence abelian. This contradicts the hypothesis that $\text{Mon}(G)$ is topologically simple and non-discrete. We conclude that $N\geqslant \text{Mon}(G)$ . Thus, every non-trivial closed normal subgroup of $H$ contains $\text{Mon}(G)$ and is thus cocompact. This proves that $G$ is hereditarily just-non-compact, as required.◻

Proof. That (ii) implies (i) is clear by observing that non-discreteness of $G^{(\infty )}$ implies that $G$ cannot be compact. We show that (i) implies (ii). Assume for contradiction that $\text{QZ}(G)$ is not discrete. Then by the assumption that $G$ is quasi just-non-compact, the subgroup $H=\overline{\text{QZ}(G)}$ is cocompact in $G$ . According to [Reference Caprace and MonodCM11, Proposition 4.3], $H$ admits a compact open normal subgroup $U$ , which by Proposition 2.3 must be contained in a compact normal subgroup of $G$ . Since every compact normal subgroup of $G$ is finite, it follows that $U$ is finite and $H$ is discrete. So, $\text{QZ}(G)$ was actually discrete.

Observe that since $G$ is quasi just-non-compact and non-discrete, an open normal subgroup of $G$ necessarily has finite index in $G$ . It follows that $G^{(\infty )}$ coincides with the intersection of all open normal subgroups of $G$ , i.e.  $G/G^{(\infty )}$ is a residually discrete group. Therefore, by Proposition 2.1, the group $G/G^{(\infty )}$ admits a basis of identity neighborhoods consisting of compact open normal subgroups. If $G^{(\infty )}$ is discrete, then it follows from Proposition 2.2 that $G$ admits compact open normal subgroups. These are necessarily finite by (i) and it follows that $G$ is discrete, which is a contradiction. So, $G^{(\infty )}$ cannot be discrete and hence $G^{(\infty )}$ is cocompact in $G$ . To obtain (ii), it only remains to observe that $\text{QZ}(G)$ cannot be cocompact, since otherwise it would contain $G^{(\infty )}$ , which would therefore be discrete, which is a contradiction.

Finally, if $N$ is a closed normal subgroup of $G$ containing $\text{QZ}(G)$ properly, then $N$ contains $G^{(\infty )}$ , so $N$ is cocompact in $G$ . Thus, $N/\text{QZ}(G)$ is cocompact in $G/\text{QZ}(G)$ , thereby confirming that $G/\text{QZ}(G)$ is just-non-compact.◻

Proof. By Proposition 2.7, the group $G$ is monolithic with non-discrete cocompact monolith $M$ , and $\text{QZ}(G)=1$ . Moreover, $\text{QZ}(M)=1$ by Proposition 2.6. In particular, $M$ is not abelian, so that $\text{C}_{G}(M)=\{1\}$ , since otherwise we would have $M\leqslant \text{C}_{G}(M)$ by definition of the monolith.

By [Reference Caprace and MonodCM11, Theorem E], the monolith $M$ has finitely many minimal closed normal subgroups $N_{1},\ldots ,N_{\ell }$ that are topologically simple, and $M=\overline{N_{1}\ldots N_{\ell }}$ . In particular, the $G$ -action by conjugation on $M$ permutes the $N_{i}$ transitively. Let $G_{0}\unlhd G$ be the open normal subgroup of finite index that normalizes $N_{i}$ for all $i$ . For every finite index open subgroup $G_{1}\leqslant G_{0}$ , we have $M\leqslant G_{1}$ , since $N_{i}\cap G_{1}$ is open of finite index in $N_{i}$ , and $N_{i}$ is topologically simple. Moreover, every non-trivial closed normal subgroup $N$ of $G_{1}$ contains one of the $N_{i}$ , since otherwise $N$ would commute with $N_{i}$ for all $i$ and hence be contained in $\text{C}_{G}(M)=\{1\}$ . Thus, $N_{1},\ldots ,N_{\ell }$ are the minimal closed normal subgroups of $G_{1}$ . It then follows from [Reference Caprace, Reid and WillisCRW17, Corollary 3.3] that $G_{1}/\text{C}_{G_{1}}(N_{i})$ is monolithic with monolith $\overline{N_{i}\text{C}_{G_{1}}(N_{i})}/\text{C}_{G_{1}}(N_{i})$ . Notice that this monolith contains $\overline{M\text{C}_{G_{1}}(N_{i})}/\text{C}_{G_{1}}(N_{i})$ and is thus cocompact. This shows that $R_{i}=G_{0}/\text{C}_{G_{0}}(N_{i})$ is hereditarily just-non-compact. It then follows from Proposition 2.7(ii) that the monolith of $R_{i}$ is non-discrete, topologically simple and cocompact. Therefore, if $R_{i}$ has a non-trivial compact locally normal subgroup that is pro- $p$ , then $R_{i}$ is pro- $p$ by Proposition 2.10.

Notice that $\bigcap _{i=1}^{\ell }\text{C}_{G_{0}}(N_{i})=\text{C}_{G_{0}}(M)\leqslant \text{C}_{G}(M)=\{1\}$ . Thus, the product homomorphism $G_{0}\rightarrow \prod _{i=1}^{\ell }R_{i}$ is injective. Now let $L$ be a non-trivial compact locally normal subgroup of $G$ that is pro- $p$ . If $L\cap G_{0}=\{1\}$ , then $L$ is finite and hence contained in $\text{QZ}(G)=\{1\}$ , which is a contradiction. Thus, we may assume without loss of generality that $L\leqslant G_{0}$ . Since the projection $G_{0}\rightarrow \prod _{i=1}^{\ell }R_{i}$ is injective, there is $i$ such that the projection of $L$ to $R_{i}$ has a non-trivial image. It then follows from the previous paragraph that $R_{i}$ is locally pro- $p$ . Since $G$ normalizes $G_{0}$ and acts transitively on the set $\{N_{1},\ldots ,N_{\ell }\}$ , the groups $R_{1},\ldots ,R_{\ell }$ are pairwise isomorphic, so that $R_{i}$ is locally pro- $p$ for every $i$ . In particular, if $V$ is a sufficiently small compact open subgroup of $G$ contained in $G_{0}$ , then the projection $G_{0}\rightarrow R_{i}$ maps $V$ onto a pro- $p$ subgroup of $R_{i}$ . Using again the injectivity of the product map $G_{0}\rightarrow \prod _{i=1}^{\ell }R_{i}$ , it follows that $V$ is pro- $p$ . This confirms that $G$ has a compact open pro- $p$ subgroup.◻

Proof. Let $V$ be the intersection of all conjugates of $U$ in $G$ . Since the reduction modulo $V$ induces a well-defined continuous surjection $\unicode[STIX]{x1D711}:\mathbf{Sub}(G)\rightarrow \mathbf{Sub}(G/V)$ such that $\unicode[STIX]{x1D711}^{-1}({\mathcal{C}}_{r,U/V}(G/V))={\mathcal{C}}_{r,U}(G)$ , it is enough to prove the statement when $V$ is trivial. In this case $G$ acts faithfully on any Cayley–Abels graph associated to $U$ and the statement follows from [Reference Caprace and RaduCR19, Proposition 2.6].◻

Proof. Let $X$ be a Cayley–Abels graph of $G$ with respect to $U$ . Then ${\mathcal{C}}_{r,U}(G)$ consists of the closed subgroups of $G$ acting with at most  $r$ orbits of vertices on $X$ . Since $X$ is connected, for every $H\in {\mathcal{C}}_{r,U}(G)$ , there is a set of representatives of the vertex orbits of $H$ that spans a connected subgraph. That subgraph has $r$ vertices and is thus of diameter  ${<}r$ .

This shows that the $r$ -ball around every vertex of $X$ contains a set of representatives of the $H$ -orbits for every $H\in {\mathcal{C}}_{r,U}(G)$ . It now suffices to choose a finite $\unicode[STIX]{x1D6F4}$ such that $\unicode[STIX]{x1D6F4}U$ contains the $r$ -ball around $U$ in $X$ .

For the converse assertion, one defines $r$ as the number of left cosets of $U$ in $\unicode[STIX]{x1D6F4}U$ and the result is clear by definition.◻

Proof. Indeed, the inclusion $O\leqslant G$ descends to an injective map of the spaces of double cosets $H\cap O\backslash O/U\cap O\rightarrow H\backslash G/U$ .◻

Proof. By Proposition 2.9, the quasi-center of $G$ is discrete. Let $U\leqslant G$ be a compact open subgroup with $U\cap \text{QZ}(G)=\{1\}$ . Let $X$ be a Cayley–Abels graph for $G$ with vertex set $G/U$ . We claim that $G$ has a Chabauty neighborhood consisting of cocompact closed subgroups $H$ with the following property: for each closed normal subgroup $N$ of $H$ , if $N\cap U\neq \{1\}$ , then $N$ is cocompact in $G$ .

If the claim fails, then there are a sequence $(H_{k})$ in $\mathbf{Sub}(G)$ converging to $G$ and a sequence $(N_{k})$ of closed normal subgroups of $H_{k}$ such that $N_{k}\cap U\neq \{1\}$ and $N_{k}$ is not cocompact in $G$ for any $k$ .

First observe that by Lemma 3.2, there is a Chabauty neighborhood of $G$ consisting of closed subgroups $H$ that are vertex-transitive on $X$ (and hence cocompact in $G$ ). Let $H\leqslant G$ be closed and vertex-transitive and let $N$ be a closed normal subgroup of $H$ with $N\cap U\neq 1$ . Let $x\in VX$ be a vertex such that $U=G_{x}$ . Then there is $r\geqslant 0$ such that $N_{x}=N_{x}^{[r]}$ and $N_{x}\not \leqslant N_{x}^{[r+1]}$ . Thus, for some vertex $y\in VX$ with $d(x,y)=r$ , we have $N_{y}\not \leqslant N_{y}^{[1]}$ . Using that $N$ is normal in $H$ and that $H$ is vertex-transitive, we deduce that $N_{x}\not \leqslant N_{x}^{[1]}$ .

Coming back to the sequence $(N_{k})$ from above, we deduce that, upon extracting, it converges to a closed normal subgroup $N$ of $G$ with $N_{x}\not \leqslant N_{x}^{[1]}$ . In particular, $N\cap U\neq \{1\}$ . Therefore, $N$ is not contained in $\text{QZ}(G)$ and it is thus cocompact in $G$ . Invoking Lemma 3.2 again, we deduce that $N_{k}$ is cocompact in $G$ for all sufficiently large $k$ , which is a contradiction. This proves the claim.

The claim directly implies that for all $H\in \mathbf{Sub}(G)$ sufficiently close to $G$ , every closed normal subgroup of $H$ that is not cocompact intersects $U$ trivially and is thus discrete. Thus, such a group $H$ is quasi just-non-compact.◻

Proof. See [Reference AschbacherAsc00, (31.4) and (31.5)]. ◻

Proof. Let $y=y_{0},y_{1},\ldots ,y_{\ell }=x$ be a shortest path from $y$ to $x$ . We have

$$\begin{eqnarray}G_{y}\unrhd G_{y}^{[1]}\unrhd G_{y,y_{1}}^{[1]}\unrhd \cdots \unrhd G_{y,y_{1},\ldots ,y_{\ell -1}}^{[1]}.\end{eqnarray}$$

Notice that $G_{x}\geqslant G_{y,y_{1},\ldots ,y_{\ell -1}}^{[1]}$ and that $G_{x}^{[r]}\leqslant G_{y,y_{1},\ldots ,y_{\ell -1}}^{[1]}$ for all $r>\ell =d(x,y)$ . Since $G_{x}^{[r]}$ is normal in $G_{x}$ , we have $G_{y,y_{1},\ldots ,y_{\ell -1}}^{[1]}\unrhd G_{x}^{[r]}$ .◻

Proof. Let $X$ be a Cayley–Abels graph for $(G,U)$ . Since $\bigcap _{g\in G}gUg^{-1}=\{1\}$ , the $G$ -action on $X$ is faithful. Since $H$ is cocompact in $G$ , it has finitely many orbits of vertices on $X$ . It follows from Lemma 3.2 that $\unicode[STIX]{x1D6E4}_{k}$ is $(r,U)$ -cocompact in $G$ for all sufficiently large $k$ , where $r=|H\backslash G/U|$ .

Let $x$ be the base vertex, i.e. the vertex corresponding to the trivial coset $U$ in $VX=G/U$ . Let $R$ be such that the ball $B(x,R)$ contains a representative of the $J$ -orbits of vertices on $X$ for all $(r,U)$ -cocompact subgroups $J$ (see Lemma 3.3).

Since $(\unicode[STIX]{x1D6E4}_{k})_{x}\neq \{1\}$ for all sufficiently large $k$ , we have $F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{x})\neq \{1\}$ , so there is $\ell _{k}\geqslant 0$ such that $F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{x})\leqslant (\unicode[STIX]{x1D6E4}_{k})_{x}^{[\ell _{k}]}$ and $F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{x})\not \leqslant (\unicode[STIX]{x1D6E4}_{k})_{x}^{[\ell _{k}+1]}$ . We claim that there exist $y_{k}$ such that $F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{y_{k}})\not \leqslant (\unicode[STIX]{x1D6E4}_{k})_{y_{k}}^{[2]}$ . If $\ell _{k}\leqslant 1$ , then we may take $y_{k}=x$ . If $\ell \geqslant 2$ , then we pick $y_{k}\in VX$ with $d(x,y_{k})=\ell _{k}-1$ and $F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{x})\not \leqslant (\unicode[STIX]{x1D6E4}_{k})_{y_{k}}^{[2]}$ . By Lemma 3.8, the group $(\unicode[STIX]{x1D6E4}_{k})_{x}^{[\ell _{k}]}$ is subnormal in $(\unicode[STIX]{x1D6E4}_{k})_{y_{k}}$ , so that $F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{x}^{[\ell _{k}]})\leqslant F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{y_{k}})$ . Since $F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{x})\leqslant (\unicode[STIX]{x1D6E4}_{k})_{x}^{[\ell _{k}]}$ , we have $F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{x})=F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{x}^{[\ell _{k}]})$ and hence $F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{x})\leqslant F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{y_{k}})$ . Therefore, $F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{y_{k}})\not \leqslant (\unicode[STIX]{x1D6E4}_{k})_{y_{k}}^{[2]}$ and we have proved the claim. Since $B(x,R)$ contains a set of representatives of the $\unicode[STIX]{x1D6E4}_{k}$ -orbits of vertices, we may assume that $y_{k}\in B(x,R)$ and hence upon extracting we may assume that $y_{k}=y$ for some $y\in B(x,R)$ and all $k$ . We set $V=G_{y}$ , which is conjugate to $U=G_{x}$ . Now we distinguish two cases.

Assume first that $E((\unicode[STIX]{x1D6E4}_{k})_{y})\not \leqslant (\unicode[STIX]{x1D6E4}_{k})_{y}^{[2]}$ for infinitely many $k$ . By definition of the layer, we may find a component $L_{k}$ of $(\unicode[STIX]{x1D6E4}_{k})_{y}$ that is not contained in $(\unicode[STIX]{x1D6E4}_{k})_{y}^{[2]}$ , and upon extracting we may assume that $(L_{k})$ converges to a subgroup $L\leqslant H_{y}$ with $L\not \leqslant H_{y}^{[2]}$ . By Lemma 3.7, the component $L_{k}$ commutes with $(\unicode[STIX]{x1D6E4}_{k})_{y}^{[2]}$ , so it follows that $L$ commutes with $H_{y}^{[2]}=\lim _{k}(\unicode[STIX]{x1D6E4}_{k})_{y}^{[2]}$ (recalling that taking an intersection with a fixed open subgroup is a Chabauty continuous operator on the $\mathbf{Sub}(G)$ ). Thus, the centralizer of $L\leqslant H_{y}$ in $H_{y}$ contains $H_{y}^{[2]}$ and is thus open. It follows that $\text{QZ}(V\cap H)=\text{QZ}(H_{y})\neq \{1\}$ . Thus, the case (i) of the statement occurs.

Assume in a second case that $E((\unicode[STIX]{x1D6E4}_{k})_{y})\leqslant (\unicode[STIX]{x1D6E4}_{k})_{y}^{[2]}$ for all but finitely many $k$ . Then we have $F((\unicode[STIX]{x1D6E4}_{k})_{y})\not \leqslant (\unicode[STIX]{x1D6E4}_{k})_{y}^{[2]}$ for all but finitely many $k$ , since $E((\unicode[STIX]{x1D6E4}_{k})_{y})F((\unicode[STIX]{x1D6E4}_{k})_{y})=F^{\ast }((\unicode[STIX]{x1D6E4}_{k})_{y})\not \leqslant (\unicode[STIX]{x1D6E4}_{k})_{y}^{[2]}$ . Since $(\unicode[STIX]{x1D6E4}_{k})_{y}$ acts faithfully on the regular graph $X$ fixing the vertex $y$ , it follows that every prime dividing the order of $(\unicode[STIX]{x1D6E4}_{k})_{y}$ is at most the degree of $X$ . Therefore, there must exist a prime $p$ such that $O_{p}((\unicode[STIX]{x1D6E4}_{k})_{y})\not \leqslant (\unicode[STIX]{x1D6E4}_{k})_{y}^{[2]}$ for infinitely many $k$ . Upon extracting, the sequence $O_{p}((\unicode[STIX]{x1D6E4}_{k})_{y})$ converges to a closed normal pro- $p$ -subgroup of $H_{y}$ that is not contained in $H_{y}^{[2]}$ . In particular, $O_{p}(H_{y})\not \leqslant \{1\}$ , so that the case (ii) of the statement occurs.◻

Proof of Proposition 3.6.

We apply Lemma 3.9. Since the conclusion (i) of the lemma is ruled out by our assumption, it follows that the conclusion (ii) must hold. Therefore, there is a conjugate $V$ of $U$ in $G$ such that $E(\unicode[STIX]{x1D6E4}_{k}\cap V)=\{1\}$ for all sufficiently large $k$ , and that $K=O_{p}(V\cap H)$ is non-trivial for some prime $p$ . Thus, $K$ is a locally normal subgroup of $H$ that is pro- $p$ . If $K$ is finite, then $K$ must lie inside $\text{QZ}(V\cap H)$ , which is absurd because $\text{QZ}(V\cap H)$ is trivial. So, $K$ is infinite and the statement holds.◻

Proof. By Proposition 2.9, the quasi-center $\text{QZ}(G)$ is discrete, so without loss of generality we may assume that $U\cap \text{QZ}(G)=\{1\}$ . Also since $\text{QZ}(G)$ is normal in $G$ and $G$ is cocompact in $L$ , upon passing to an open subgroup of $U$ we may also assume that $\text{QZ}(G)$ intersects trivially all $L$ -conjugates of $U$ (by replacing $U$ by $\bigcap _{i=1}^{r}l_{i}Ul_{i}^{-1}$ , where $l_{1}U,\ldots ,l_{r}U$ are representatives for the $G$ -orbits in $L/U$ ). Hence, whenever $V$ is an $L$ -conjugate of $U$ , we have $\text{QZ}(V\cap H)\leqslant V\cap \text{QZ}(G)=1$ , since $H$ is an open subgroup of $G$ . Since $H$ is non-discrete, the discrete groups $\unicode[STIX]{x1D6E4}_{k}$ intersect $U$ non-trivially for large enough $k$ . Therefore, it follows from Proposition 3.6 that there is a prime $p$ such that $H$ has an infinite compact locally normal subgroup $K$ that is pro- $p$ . Since $H$ is open in $G$ , that subgroup $K$ is also locally normal in $G$ . Note that $(K\cap U)\cap \text{QZ}(G)\leqslant U\cap \text{QZ}(G)=1$ , so the quotient group $G/\text{QZ}(G)$ also has an infinite compact subgroup that is locally normal and pro- $p$ . By Proposition 2.9,  $G/\text{QZ}(G)$ is just-non-compact, and it is also compactly generated, so by Proposition 2.11 the group $G/\text{QZ}(G)$ must have an open pro- $p$ subgroup. It follows that the same is true in $G$ , since $\text{QZ}(G)$ is a discrete subgroup of $G$ .◻

Proof of Theorem F.

We let $G$ be a non-discrete, compactly generated, quasi just-non-compact tdlc group, and $H$ an open subgroup of finite index in $G$ that is a Chabauty limit of a sequence of discrete subgroups of $G$ . Since $G$ is quasi just-non-compact, every compact normal subgroup of $G$ is finite. In particular, the group $G$ admits a compact open subgroup $U$ such that $\bigcap _{g\in G}gUg^{-1}=\{1\}$ . Hence, we may apply Theorem 3.10 with $L=G$ , and we deduce that there exist a prime $p$ and an open pro- $p$ subgroup $V$ of $G$ . We have to show that $V$ is not topologically finitely generated. We argue by contradiction and assume that $V$ is topologically finitely generated. Then so is $W=V\cap H$ . Since $W$ is also pro- $p$ , it follows that $W$ has an open Frattini subgroup [Reference Dixon, du Sautoy, Mann and SegalDSMS91, Proposition 1.14] and hence $W$ is an isolated point of $\mathbf{Sub}(W)$ [Reference Gartside and SmithGS10, Theorem 5.6]. Since $(\unicode[STIX]{x1D6E4}_{k})$ converges to $H$ , the sequence of intersections $(\unicode[STIX]{x1D6E4}_{k}\cap W)$ converges to $W$ , so that $\unicode[STIX]{x1D6E4}_{k}$ contains the open subgroup $W$ for all sufficiently large $k$ , contradicting that $\unicode[STIX]{x1D6E4}_{k}$ is discrete.◻

Proof. Let $U_{1}$ be a compact open subgroup of $G_{1}$ and let $O=H\cap (U_{1}\times G_{2})$ . Note that $O$ is compactly generated as it is cocompact in the compactly generated group $U_{1}\times G_{2}$ . We write $N=H\,\cap \,G_{2}$ , which is contained in $O$ . The group $O/N$ embeds continuously in the profinite group $U_{1}$ and hence $O/N$ is residually discrete. Since $N$ is discrete by assumption, Proposition 2.1 together with Proposition 2.2 imply that the group $O$ admits a compact open normal subgroup $K$ . Now according to Proposition 2.3 the subgroup $K$ is contained in a compact normal subgroup of $U_{1}\times G_{2}$ , since $K$ has a cocompact normalizer in $U_{1}\times G_{2}$ . But $G_{2}$ has a discrete polycompact radical, so every compact normal subgroup of $U_{1}\times G_{2}$ has a finite projection to the factor $G_{2}$ . In particular, there is a finite index closed subgroup of $K$ that is contained in $U_{1}$ , and $H\cap G_{1}$ is open in $H$ .◻

Proof. The proof follows the same lines as Proposition 4.1. The subgroup $O=H\cap (U_{1}\times G_{2})$ has finite covolume in $U_{1}\times G_{2}$ and hence also has property (T). In particular, $O$ is compactly generated. By the same argument as above, we see that $O$ admits a compact open normal subgroup $K$ , so that $p_{2}(K)$ is a compact normal subgroup of $p_{2}(O)$ . Now since $p_{2}(O)$ has finite covolume in $G_{2}$ , the amenable radical of $p_{2}(O)$ must be contained in the amenable radical of $G_{2}$ (by [Reference FurstenbergFur81, Proposition 4.4] and [Reference FurmanFur03, Proposition 7]; or by the main result of [Reference Bader, Duchesne and LécureuxBDL16]). Since $G_{2}$ has a discrete amenable radical, the subgroup $p_{2}(K)$ is therefore contained in a discrete subgroup of $G_{2}$ . So, $K$ has a finite projection to $G_{2}$ , and $H\cap G_{1}$ is open in $H$ .◻

Proof. Since the polycompact radical of a finite product is the product of the polycompact radicals, every subproduct of $G$ has a discrete polycompact radical, so we deduce that $\text{Ker}(p_{i}|_{H})$ is open in $H$ for every $i$ according to Proposition 4.1. Therefore, $\bigcap _{i}\text{Ker}(p_{i}|_{H})=1$ remains open in $H$ and hence $H$ is discrete.◻

Proof. For each compact open subgroup $W\leqslant G_{1}$ , the intersection $H\cap (W\times G_{2})$ contains $H\cap G_{2}$ . Thus, the number of orbits on $G_{2}/U_{2}$ of the projection $p_{2}(H\cap (W\times G_{2}))$ is bounded above by the number of $(H\cap G_{2})$ -orbits, which is finite in view of the hypothesis that $H\cap G_{2}$ is cocompact in $G_{2}$ . Let $M$ be the maximum of the number of orbits over all possible $W$ , and let $U_{1}\leqslant G_{1}$ be a compact open subgroup such that $p_{2}(H\cap (U_{1}\times G_{2}))$ has exactly $M$ orbits on $G_{2}/U_{2}$ . Then, for every open subgroup $V_{1}\leqslant U_{1}$ , the projection $p_{2}(H\cap (V_{1}\times G_{2}))$ has also $M$ orbits on $G_{2}/U_{2}$ . Since $H\cap (V_{1}\times G_{2})$ is a subgroup of $H\cap (U_{1}\times G_{2})$ , it follows that $p_{2}(H\cap (V_{1}\times G_{2}))$ and $p_{2}(H\cap (U_{1}\times G_{2}))$ must have the same orbits on $G_{2}/U_{2}$ .◻

Proof. (i) We fix a compact open subgroup $U_{1}\leqslant G_{1}$ afforded by Lemma 4.4. Also let $u\in U_{1}$ and $V_{1}\leqslant U_{1}$ be any open subgroup of $U_{1}$ . Since $p_{1}(H)$ is dense in $G_{1}$ , there exists $h\in H$ with $p_{1}(h)\in V_{1}u\subset U_{1}$ . In particular, we have $h\in H\cap (U_{1}\times G_{2})$ . By Lemma 4.4, there exists $h^{\prime }\in H\cap (V_{1}\times G_{2})$ such that $h^{\prime \prime }=h^{\prime }h\in G_{1}\times U_{2}$ . We have $h^{\prime \prime }\in H\cap (G_{1}\times U_{2})$ and $p_{1}(h^{\prime \prime })=p_{1}(h^{\prime })p_{1}(h)\in V_{1}u$ . Since $V_{1}$ was arbitrary, we infer that the closure of $p_{1}(H\cap (G_{1}\times U_{2}))$ contains $U_{1}$ and is thus open in $G_{1}$ . But the subgroup $U_{2}$ being compact, the projection $p_{1}(H\cap (G_{1}\times U_{2}))$ is closed, so we actually deduce that $p_{1}(H\cap (G_{1}\times U_{2}))$ is open, as required.

(ii) By hypothesis, $p_{1}(H)$ is dense in $G_{1}$ and, by (i), $p_{1}(H)$ is open in $G_{1}$ . Thus, $p_{1}(H)=G_{1}$ . Now let $g\in G_{1}$ and choose $h\in H$ with $p_{1}(h)=g$ . Since $(H\cap G_{2})U_{2}=G_{2}$ , there exists $h^{\prime }\in H\cap G_{2}$ with $h^{\prime }p_{2}(h)=p_{2}(h^{\prime }h)\in U_{2}$ . Hence, $h^{\prime }h\in H\cap (G_{1}\times U_{2})$ . Since $p_{1}(h^{\prime })=1$ , we have $p_{1}(h^{\prime }h)=p_{1}(h)=g$ .◻

Proof. For every $i$ , we write $K_{i}=H\cap G_{i}$ . The subgroup $K_{i}$ is centralized by $\prod _{j\neq i}G_{j}$ and normalized by $H$ . So, the normalizer of $K_{i}$ in $G$ , which is a closed subgroup, contains the subgroup $(\prod _{j\neq i}G_{j})H$ , which is dense in $G$ by assumption. Therefore, $K_{i}$ is normal in $G$ and hence is either discrete or cocompact in $G_{i}$ , since $G_{i}$ is quasi just-non-compact.

We let $\unicode[STIX]{x1D6F1}\subseteq \{1,\ldots ,n\}$ be the set of $i$ such that $K_{i}$ is not discrete, and $\unicode[STIX]{x1D6F4}$ be the complement of $\unicode[STIX]{x1D6F1}$ . We have to show that $\unicode[STIX]{x1D6F1}=\{1,\ldots ,n\}$ and $\unicode[STIX]{x1D6F4}=\emptyset$ . The group $K_{\unicode[STIX]{x1D6F1}}=\prod _{i\in \unicode[STIX]{x1D6F1}}K_{i}$ is cocompact in $G_{\unicode[STIX]{x1D6F1}}$ , so the projection $p_{\unicode[STIX]{x1D6F4}}:G\rightarrow G_{\unicode[STIX]{x1D6F4}}$ factors through a proper map $G/K_{\unicode[STIX]{x1D6F1}}\rightarrow G_{\unicode[STIX]{x1D6F4}}$ . Moreover, $K_{\unicode[STIX]{x1D6F1}}$ is contained in $H$ , so that the quotient $H/K_{\unicode[STIX]{x1D6F1}}$ is a closed subgroup of $G/K_{\unicode[STIX]{x1D6F1}}$ . Since the map $G/K_{\unicode[STIX]{x1D6F1}}\rightarrow G_{\unicode[STIX]{x1D6F4}}$ is proper, it follows that the projection of $H/K_{\unicode[STIX]{x1D6F1}}$ to $G_{\unicode[STIX]{x1D6F4}}$ , which is also the projection of $H$ to $G_{\unicode[STIX]{x1D6F4}}$ , has a closed image.

If (1) holds, then we may invoke Proposition 4.1 and deduce that $\text{Ker}(p_{j}|_{H})$ is open in $H$ for every $j\notin \unicode[STIX]{x1D6F1}$ . If (2) holds, then we observe that for every $i$ , the amenable radical of $G_{i}$ is discrete, since it cannot be cocompact by the hypothesis that $G_{i}$ is a non-compact group with property (T). In particular, the amenable radical of $\prod _{i\neq j}G_{i}$ is discrete for all $j$ . Therefore, by Proposition 4.2, the conclusion of Proposition 4.1 also holds in that case and we deduce again that for every $j\notin \unicode[STIX]{x1D6F1}$ , the subgroup $\text{Ker}(p_{j}|_{H})$ is open in $H$ .

It follows that $H\cap G_{\unicode[STIX]{x1D6F1}}=\bigcap _{j\notin \unicode[STIX]{x1D6F1}}\text{Ker}(p_{j}|_{H})$ is an open subgroup of $H$ . Since the projection $H\rightarrow G_{\unicode[STIX]{x1D6F4}}$ has closed image and factors through the discrete group $H/H\cap G_{\unicode[STIX]{x1D6F1}}$ , we infer that the projection of $H$ to $G_{\unicode[STIX]{x1D6F4}}$ has discrete image. In view of the hypotheses, this implies that $\unicode[STIX]{x1D6F4}$ is the empty set. Therefore, $\unicode[STIX]{x1D6F1}=\{1,\ldots ,n\}$ and the proof of (i) is complete.

In order to prove (ii), we fix $i$ and write $L_{i}=p_{i}(H\cap (U_{1}\times \cdots \times G_{i}\times \cdots \times U_{n}))$ . According to (i), we have that $H\,\cap \,\prod _{j\neq i}G_{j}$ is cocompact in $\prod _{j\neq i}G_{j}$ , so we may apply Proposition 4.5, which says that $L_{i}$ is an open subgroup of $G_{i}$ . But by (i) again we also know that $L_{i}$ is cocompact in $G_{i}$ and hence $L_{i}$ is a finite index subgroup of $G_{i}$ .◻

Proof. First note that the condition (Irr3) implies that all factors $G_{i}$ are non-discrete. Fix $\unicode[STIX]{x1D6F4}\subsetneq \{1,\ldots ,n\}$ and write $H=\overline{p_{\unicode[STIX]{x1D6F4}}(\unicode[STIX]{x1D6E4})}$ . Note that there is nothing to prove if $\unicode[STIX]{x1D6F4}$ is a singleton, so we assume that $\unicode[STIX]{x1D6F4}$ has cardinality at least two. The subgroup $H$ is closed and cocompact in $G_{\unicode[STIX]{x1D6F4}}$ and has a dense projection on each $G_{i}$ for $i\in \unicode[STIX]{x1D6F4}$ . Since $\unicode[STIX]{x1D6E4}$ satisfies (Irr3), Proposition 4.6 can be applied to the group $H$ inside $G_{\unicode[STIX]{x1D6F4}}$ . So, we deduce that $H$ contains $\prod _{i\in \unicode[STIX]{x1D6F4}}K_{i}$ for some closed cocompact normal subgroups $K_{i}\leqslant G_{i}$ . By Proposition 2.9, we must have $G_{i}^{(\infty )}\leqslant K_{i}$ , whence the statement.◻

Proof. We let $H$ be the closure of $\operatorname{Comm}_{G}(\unicode[STIX]{x1D6E4})$ in $G$ . We have to show that either $H$ is discrete or $H=G$ . Assume that $H$ is not discrete. Then we may apply Proposition 4.6 and we deduce that $H$ contains $K_{1}\times \cdots \times K_{n}$ for some closed cocompact normal subgroups $K_{i}\lhd G_{i}$ . That $G_{i}=G_{i}^{\infty }$ is equivalent to saying that $G_{i}$ has no proper cocompact closed normal subgroup, so it follows that $H$ contains $G_{i}$ for all but possibly one factor. But since $H$ is a closed subgroup of $G$ and $H$ has a dense projection to this remaining factor, we deduce that $H=G$ , as required.◻

Proof. If (Irr3) fails, there exists a partition $\unicode[STIX]{x1D6F4}\cup \unicode[STIX]{x1D6F1}=\{1,\ldots ,n\}$ with $\unicode[STIX]{x1D6F1},\unicode[STIX]{x1D6F4}\neq \varnothing$ such that $p_{\unicode[STIX]{x1D6F4}}(\unicode[STIX]{x1D6E4})$ is discrete in $G_{\unicode[STIX]{x1D6F4}}$ . Since $\overline{p_{\unicode[STIX]{x1D6F4}}(\unicode[STIX]{x1D6E4})}$ is also a lattice in $G_{\unicode[STIX]{x1D6F4}}$ , it follows that $p_{\unicode[STIX]{x1D6F4}}$ maps $\unicode[STIX]{x1D6E4}$ onto a lattice in $G_{\unicode[STIX]{x1D6F4}}$ . It then follows from [Reference RaghunathanRag72, Theorem I.1.13] that $\text{Ker}(p_{\unicode[STIX]{x1D6F4}}|_{\unicode[STIX]{x1D6E4}})=\unicode[STIX]{x1D6E4}\cap G_{\unicode[STIX]{x1D6F1}}$ is a lattice in $G_{\unicode[STIX]{x1D6F1}}$ , which is non-compact by assumption. So, $\unicode[STIX]{x1D6E4}\cap G_{\unicode[STIX]{x1D6F1}}$ is non-trivial, which contradicts (Irr2).

If (Irr0) fails, i.e. if $(G_{\unicode[STIX]{x1D6F1}}\,\cap \,\unicode[STIX]{x1D6E4})(G_{\unicode[STIX]{x1D6F4}}\,\cap \,\unicode[STIX]{x1D6E4})$ is of finite index in $\unicode[STIX]{x1D6E4}$ for some partition $\unicode[STIX]{x1D6F1}\cup \unicode[STIX]{x1D6F4}=\{1,\ldots ,n\}$ with $\unicode[STIX]{x1D6F1},\unicode[STIX]{x1D6F4}\neq \varnothing$ , then arguing as above we see that $G_{\unicode[STIX]{x1D6F1}}\cap \unicode[STIX]{x1D6E4}$ is a lattice in $G_{\unicode[STIX]{x1D6F1}}$ and obtain again a contradiction with (Irr2).

If (Irr0) fails, then there exists a partition $\unicode[STIX]{x1D6F1}\cup \unicode[STIX]{x1D6F4}=\{1,\ldots ,n\}$ with $\unicode[STIX]{x1D6F1},\unicode[STIX]{x1D6F4}\neq \varnothing$ such that the direct product $\unicode[STIX]{x1D6E4}_{1}=(G_{\unicode[STIX]{x1D6F1}}\cap \unicode[STIX]{x1D6E4})(G_{\unicode[STIX]{x1D6F4}}\cap \unicode[STIX]{x1D6E4})$ is of finite index in $\unicode[STIX]{x1D6E4}$ . In particular, $\unicode[STIX]{x1D6E4}_{1}$ is a lattice in $G=G_{\unicode[STIX]{x1D6F1}}\times G_{\unicode[STIX]{x1D6F4}}$ . It follows that $G_{\unicode[STIX]{x1D6F1}}\cap \unicode[STIX]{x1D6E4}$ is a lattice in $G_{\unicode[STIX]{x1D6F1}}$ and $G_{\unicode[STIX]{x1D6F4}}\cap \unicode[STIX]{x1D6E4}$ is a lattice in $G_{\unicode[STIX]{x1D6F4}}$ . Thus, both factors in the direct product decomposition of $\unicode[STIX]{x1D6E4}_{1}$ are non-trivial, so that (Irr4) fails.◻