Conjugating automorphisms of graph products: Kazhdan's property (T) and SQ-universality

TL;DR

This paper characterizes when the group of conjugating automorphisms of a graph product of groups has Kazhdan's property (T) and exhibits SQ-universality, revealing deep structural properties of these automorphism groups.

Contribution

It provides a precise characterization of conditions under which conjugating automorphism groups of graph products have property (T) and are SQ-universal.

Findings

Conjugating automorphism groups satisfy property (T) under specific conditions.

They are SQ-universal in certain cases.

The paper offers a complete classification for these properties.

Abstract

An automorphism of a graph product of groups is conjugating if it sends each factor to a conjugate of a factor (possibly different). In this article, we determine precisely when the group of conjugating automorphisms of a graph product satisfies Kazhdan's property (T) and when it satisfies some vastness properties including SQ-universality.

Equations

$$\left(\underset{u\in V(\Gamma )}{\ast }{G}_u\right)/⟨⟨[g,h]=1,g\in G_u,h\in G_v,\{u,v\}\in E(\Gamma )⟩⟩$$

$$|\phi (gh)-\phi (g)-\phi (h)|\le C\text{for every g,h∈G}.$$

$$\psi \mapsto \left(v\mapsto \text{vertex whose stabiliser is ψ(stab(v))}\right)$$

$$g:=s_1{r}_1\cdot s_2{r}_2\cdots s_n{r}_n\in A\ast B.$$

$$A,s_1\cdot B,s_1{r}_1\cdot A,s_1{r}_1{s}_2\cdot B,s_1{r}_1{s}_2{r}_2\cdot A,\dots ,s_1{r}_1\cdots s_{n-1}{r}_{n-1}{s}_n{r}_n\cdot A=g\cdot A$$

$$s_1{r}_1\cdots s_{n-1}{r}_{n-1}{s}_n\cdot B=s_1{r}_1\cdots s_{n-1}{r}_{n-1}{s}_n{r}_n\cdot B=g\cdot B,$$

$$\mathrm{stab}(A)\cap \mathrm{stab}(B)\cap \mathrm{stab}(g\cdot A)\cap \mathrm{stab}(g\cdot B)$$

$$F:=\{\phi \in \mathrm{PConjAut}(A\ast B)\mid \phi (A)=A,\phi (B)=B,\phi (g\cdot A)=g\cdot A,\phi (g\cdot B)=g\cdot B\}$$

$$s_1{r}_1\cdot s_2{r}_2\cdots s_n{r}_n=g=\phi (g)=\phi (s_1)\phi (r_1)\cdot \phi (s_2)\phi (r_2)\cdots \phi (s_n)\phi (r_n)$$

$$\mathrm{PConjAut}(\Gamma \mathcal{G})=\underset{u\in V(\Gamma )}{⨁}\mathrm{Inn}(G_u)\simeq \underset{u\in V(\Gamma )}{⨁}{G}_u/Z(G_u).$$

$$\mathrm{PConjAut}(C_{\Xi })\oplus \underset{u\in V(\Xi )}{⨁}\mathrm{Inn}(G_u)\simeq \mathrm{PConjAut}(C_{\Xi })\oplus \underset{u\in V(\Xi )}{⨁}{G}_u/Z(G_u).$$

Full text

A. G. was supported by a public grant as part of the Fondation Mathématique Jacques Hadamard.

O. V. was supported by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany’s Excellence Strategy -EXC 2044-, Mathematics Münster: Dynamics-Geometry-Structure

Given a simplicial graph $\Gamma$ and a collection of groups $\mathcal{G}=\{G_{u}\mid u\in V(\Gamma)\}$ indexed by the vertex-set $V(\Gamma)$ of $\Gamma$, the graph product $\Gamma\mathcal{G}$ is defined as the quotient

[TABLE]

where $E(\Gamma)$ denotes the edge-set of $\Gamma$. Two cases to keep in mind is that the graph product $\Gamma\mathcal{G}$ reduces to the free product of $\mathcal{G}$ if $\Gamma$ does not contain any edge, and it reduces to the direct sum of $\mathcal{G}$ if any two vertices of $\Gamma$ are linked by an edge. Usually, one says that graph products interpolate between free products and direct sums. In particular, they provide a convenient framework in which studying right-angled Artin groups (when factors are infinite cyclic) and right-angled Coxeter groups (when factors are cyclic of order two) in a unified way.

In this article, we are interested in conjugating automorphisms of graph products, ie., automorphisms sending factors to conjugates of factors. Groups of conjugating automorphisms of (particular) graph products turn out to appear in different places in the literature. For free groups, these automorphism groups have a very nice topological interpretation, as they coincide with motion groups of circles in the Euclidean spaces [10] (see also [5] and references therein). Conjugating automorphisms also appear naturally when studying free products [7, 8], and have been studied independently [19]. In addition to their own interest, studying conjugating automorphisms have been a key step in the understanding of the global automorphism group in several places; see for instance [26] for right-angled Coxeter groups and [16] for right-angled Artin groups. Also, the group of conjugating automorphisms turns out to coincides with (a finite-index subgroup of) the global automorphism group in some cases. For instance, as proved in [9], the group of conjugating automorphisms of a graph product of finite groups has finite index in the entire automorphism group.

Given a graph product $\Gamma\mathcal{G}$, we denote by $\mathrm{ConjAut}(\Gamma\mathcal{G})$ the group of its conjugating automorphisms, and by $\mathrm{PConjAut}(\Gamma\mathcal{G})$ its subgroup whose elements send each factor to a conjugate of itself. Also, we emphasize that, in all the article, we use the following convention: factors of graph products (also referred to as vertex-groups) are always supposed to be non-trivial.

The main result of our article is the following:

Recall from [20] that, if a group $G$ acts on a simplicial tree $T$, then $g\in G$ is WPD (for Weakly Properly Discontinuous) if it is loxodromic and if its axis contains two vertices $u$ and $v$ (not necessarily distinct) such that $\mathrm{stab}_{G}(u)\cap\mathrm{stab}_{G}(v)$ is finite.

We deduce many interesting consequences from our theorem. First, we deduce some information regarding Kazhdan’s Property (T). Examples of groups with Kazhdan’s property (T) are the general linear groups ${\rm GL}_{n}(\mathbb{Z})$ for $n\geq 3$ ([2, 4.2.5]) and the automorphism groups of free groups ${\rm Aut}(F_{n})$ for $n\geq 5$ ([14], [15]). More generally, an interesting question is the following: For which shape of the graph $\Gamma$ does the automorphism group ${\rm Aut}(\Gamma\mathcal{G})$ have Kazhdan’s property (T)? There are some results in the literature regarding (outer) automorphism groups of graph products of finite abelian groups in [17, Thm. 1.4], [24, Cor. 3], [27, Thm. C], of right-angled Artin groups in [1, Cor. 1.8] and of graph products of arbitrary vertex-groups in [9, Thm. F].

Our main result in this direction is:

Because Property (T) is stable under commensurability [2, Theorem 1.7.1] and because the group of conjugating automorphisms of a graph product of finite groups has finite-index in the entire automorphism group [9], we get that:

A good reference for the above corollary is [18]. We also deduce from the combination of Theorem A with [9, Theorem 3.23] the following statement, which generalises [9, Theorem F]:

Recall from [9] that a graph is molecular if it is finite, connected, leafless and if it has girth $\geq 5$ (ie., it is triangle- and square-free).

Our second main application deals with SQ-universality. Recall that a group $G$ is SQ-universal if, for every countable group $Q$, there exists $N\lhd G$ such that $Q$ embeds into $G/N$. Interestingly, a SQ-universal group has to contain a non-abelian free subgroup and uncountably many normal subgroups. The last observation motivates the idea that being SQ-universal can be thought of as a strong negation of being simple. The first known examples of SQ-universal groups are free groups, so that large groups (ie., groups containing finite-index subgroups which surject onto a non-abelian free subgroup) are also SQ-universal. Now, thanks to the theory of acylindrically hyperbolic groups, we know that many groups turn out to be SQ-universal; see [22] and references therein for more information.

Actually, the strategy used to prove Theorem B can be also applied to other vastness properties, including virtually having an infinite-dimensional space of quasimorphisms and not being boundedly generated.

Recall that, given a group $G$, a quasimorphism is a map $\varphi:G\to\mathbb{R}$ such that there exists some constant $C\geq 0$ so that

[TABLE]

Such a quasimorphism is homogeneous if $\varphi(g^{k})=k\cdot\varphi(g)$ for every $g\in G$ and $k\in\mathbb{Z}$. Notice that the set of all the homogeneous quasimorphisms of our group $G$ is naturally a $\mathbb{R}$-vector space, which we denote by $\mathrm{QH}(G)$. Following [11], we say that $G$ virtually has many quasimorphisms if it contains a finite-index subgroup $H$ such that $\mathrm{QH}(H)$ is infinite-dimensional.

Next, recall that a group $G$ is boundedly generated if there exist finitely many elements $g_{1},\ldots,g_{n}\in G$ such that, for every $g\in G$, there exist integers $m_{1},\ldots,m_{n}\in\mathbb{Z}$ so that $g=g_{1}^{m_{1}}\cdots g_{n}^{m_{n}}$.

Because the group of conjugating automorphisms of a graph product of finite groups turns out to have finite-index in the entire automorphism group [9], we deduce the following statement from the theorems above:

1. (i)

The subgraph $\Xi$ contains a vertex which is not labelled by $\mathbb{Z}_{2}$ or $\Xi$ does not decompose as the join of a complete graph with copies of the disjoint union of two single vertices. 2. (ii)

The automorphism group $\mathrm{Aut}(\Gamma\mathcal{G})$ involves all finite groups. 3. (iii)

The automorphism group $\mathrm{Aut}(\Gamma\mathcal{G})$ is SQ-universal. 4. (iv)

The automorphism group $\mathrm{Aut}(\Gamma\mathcal{G})$ virtually has many quasimorphisms. 5. (v)

The automorphism group $\mathrm{Aut}(\Gamma\mathcal{G})$ is not boundedly generated. 6. (vi)

The group $\Gamma\mathcal{G}$ is not virtually free abelian.

In this statement, following [11], we say that a group $G$ involves all finite groups if any finite group appears as a quotient of a finite-index subgroup of $G$.

Proof of the main theorem

Let $\Gamma$ be a simplicial graph and $\mathcal{G}$ a collection of groups indexed by $V(\Gamma)$. Assume that $\Gamma$ is not complete.

Fix two vertices $u,v\in V(\Gamma)$ which are not adjacent. Notice that, as $\mathrm{PConjAut}(\Gamma\mathcal{G})$ sends each vertex-group to a conjugate of itself, the normal closure $Q$ in $\Gamma\mathcal{G}$ of the subgroup $\langle G_{w},\ w\in V(\Gamma)\backslash\{u,v\}\rangle$ is preserved by $\mathrm{PConjAut}(\Gamma\mathcal{G})$. Moreover, the inclusion $\langle G_{u},G_{v}\rangle\hookrightarrow\Gamma\mathcal{G}$ induces an isomorphism $\langle G_{u},G_{v}\rangle\to\Gamma\mathcal{G}/Q$, and the subgroup $\langle G_{u},G_{v}\rangle$ is naturally isomorphic to the free product $G_{u}\ast G_{v}$. As a consequence, the quotient $\Gamma\mathcal{G}\to\Gamma\mathcal{G}/Q$ induces a homomorphism $\Phi:\mathrm{PConjAut}(\Gamma\mathcal{G})\to\mathrm{PConjAut}(G_{u}\ast G_{v})$. Notice that $\Phi(\mathrm{Inn}(\Gamma\mathcal{G}))=\mathrm{Inn}(G_{u}\ast G_{v})$, so that $\mathrm{Inn}(G_{u}\ast G_{v})$ is included into the image of $\Phi$.

The fact that $\mathrm{PConjAut}(\Gamma\mathcal{G})$ acts fixed-point freely on a simplicial tree now follows from the following observation:

Claim**.**

Let $A,B$ be two non-trivial groups. If $T$ denotes the Bass-Serre tree of the free product $A\ast B$, then

[TABLE]

defines an isometric action $\mathrm{PConjAut}(A\ast B)\curvearrowright T$ which extends $A\ast B\curvearrowright T$ when $A\ast B$ is canonically identified to $\mathrm{Inn}(A\ast B)$.

We refer to [25, §4, Thm. 7] for more information on Bass-Serre trees. A proof of the previous claim can be found in [13] and in [23, Thm. 1] in greater generality. We include here the sketch of a direct proof for reader’s convenience.

Sketch of proof of the claim..

The vertices of the Bass-Serre tree $T$ are the cosets of $A$ and $B$. Because $A$ and $B$ are malnormal subgroups of $A\ast B$ (ie., for every $g\in A\ast B$, the intersection $A\cap gBg^{-1}$ is always trivial and the intersection $A\cap gAg^{-1}$ is non-trivial only when $g\in A$), there is a natural bijection between the cosets of $A$ and $B$ and their conjugates. By definition of $\mathrm{PConjAut}(\Gamma\mathcal{G})$, it follows that

[TABLE]

defines an action of $\mathrm{PConjAut}(\Gamma\mathcal{G}))$ on the vertices of $T$. It remains to show that the action preserves the adjacency relation of the vertices of $T$.

Notice that, if $u,v\in T$ are two vertices, then the geodesic $[u,v]$ is a fundamental domain for the action of the subgroup $\langle\mathrm{stab}(u),\mathrm{stab}(v)\rangle\leq A\ast B$ on $T$. Therefore, since $A\ast B$ acts on $T$ with a single orbit of edges, the two vertices $u$ and $v$ turn out to be adjacent if and only if $\langle\mathrm{stab}(u),\mathrm{stab}(v)\rangle=A\ast B$. Because this characterisation is purely algebraic, it must be preserved by automorphisms of $A\ast B$, concluding the proof. ∎

Thus, we have proved the first assertion of our theorem. The second assertion now follows from the following observation:

Claim**.**

Let $A,B$ be two finitely generated groups. Then the action of $\mathrm{PConjAut}(A\ast B)$ on the Bass-Serre tree $T$ of $A\ast B$ contains a WPD isometry.

Fix two generating sets $\{s_{1},\ldots,s_{n}\}$ and $\{r_{1},\ldots,r_{n}\}$ of $A$ and $B$ respectively, such that $r_{i}$ and $s_{i}$ are non-trivial for every $1\leq i\leq n$. Set

[TABLE]

We claim that the inner automorphism $\iota(g)\in\mathrm{PConjAut}(A\ast B)$ defines a WPD isometry of $T$. First of all, notice that, if $\sigma\subset T$ denotes the segment

[TABLE]

then $\gamma:=\bigcup\limits_{k\in\mathbb{Z}}g^{k}\sigma$ defines a geodesic on which $g$ acts by translations. Notice also that the penultimate vertex of $\sigma$ is

[TABLE]

so that $B$ belongs to $\gamma$. Therefore, $\gamma$ is an axis of $\iota(h)$ which contains the vertices $A$, $B$, $g\cdot A$ and $g\cdot B$. As a consequence of [20, Corollary 4.3], in order to show that $\iota(h)$ is WPD, it is sufficient to verify that the intersection

[TABLE]

is finite, where the stabilisers are taken with respect to the action of $\mathrm{PConjAut}(A\ast B)$. Otherwise saying, we want to show that

[TABLE]

is finite. So let $\varphi\in F$. Because $g\cdot A=\varphi(g\cdot A)=\varphi(g)\cdot A$ and $g\cdot B=\varphi(g\cdot B)=\varphi(g)\cdot B$, we deduce that $g^{-1}\varphi(g)\in A\cap B=\{1\}$, hence $\varphi(g)=g$. Because $\varphi$ stabilises both $A$ and $B$, it follows that

[TABLE]

is an equality between two alternating words, so we must have $\varphi(s_{i})=s_{i}$ and $\varphi(r_{i})=r_{i}$ for every $1\leq i\leq n$. But $\{s_{1},\ldots,s_{n}\}$ and $\{r_{1},\ldots,r_{n}\}$ generate $A$ and $B$ respectively, so $\varphi$ must be the identity. Thus, we have proved that $F=\{1\}$, which concludes the proof of our claim.

Remark 1**.**

It is worth noticing that the assumption on the finite generation of vertex-groups can be weakened. Indeed, we only need each vertex-group to contain a finite set such that any automorphism fixing pointwise this set has to be the identity. Of course, taking a finite generating set works, but groups which are not finitely generated may also contain such sets. For instance, any automorphism of $\mathbb{Q}$ fixing $\{1\}$ must be the identity.

Proofs of Theorems A, B, C and D

We begin by proving Theorem A about Kazhdan’s Property (T).

Proof of Theorem A.

If $\Gamma$ is not complete, then it follows from our main theorem that $\mathrm{PConjAut}(\Gamma\mathcal{G})$ acts fixed-point freely on a simplicial tree. Because Property (T) is stable under commensurability [2, Theorem 1.7.1] and that a group acting fixed-point freely on a simplicial tree cannot satisfy Property (T) [28], we conclude that $\mathrm{ConjAut}(\Gamma\mathcal{G})$ does not satisfy Property (T).

Conversely, if $\Gamma$ is a complete graph, then

[TABLE]

Because Property (T) is stable under commensurability and because a direct sum satisfies Property (T) if and only if so do its factors [2, Proposition 1.7.8], we conclude that $\mathrm{ConjAut}(\Gamma\mathcal{G})$ satisfies Property (T) if and only if so does $G_{u}/Z(G_{u})$ for every $u\in V(\Gamma)$. ∎

Next, let us turn to the proof of Theorem B and its corollaries. From now on, we fix a group property $\mathcal{P}$ satisfying the following conditions:

• (i)

acylindrically hyperbolic groups satisfy $\mathcal{P}$;

• (ii)

if a group has a quotient satisfying $\mathcal{P}$, then it satisfies $\mathcal{P}$ as well;

• (iii)

a group satisfies $\mathcal{P}$ if and only if so do its finite-index subgroups;

• (iv)

the direct sum of two groups satisfies $\mathcal{P}$ if and only if so does one of the two factors;

• (v)

free abelian groups do not satisfy $\mathcal{P}$.

We refer to [22] for more information on acylindrically hyperbolic groups. The only thing we need to know that, if a group which is not virtually cyclic acts on a simplicial tree with a WPD isometry, then it has to be acylindrically hyperbolic.

Our goal now is to show the following statement:

Theorem F**.**

Let $\Gamma$ be a finite simplicial graph and $\mathcal{G}$ a collection of finitely generated groups indexed by $V(\Gamma)$. Decompose $\Gamma$ as a join $\Lambda\ast\Xi$ where $\Lambda$ is the subgraph generated by the vertices which are adjacent to all the vertices of $\Gamma$. Then $\mathrm{ConjAut}(\Gamma\mathcal{G})$ satisfies $\mathcal{P}$ if and only if $\Xi$ contains a vertex which is not labelled by $\mathbb{Z}_{2}$, or $\Lambda$ contains a vertex $u$ such that $G_{u}/Z(G_{u})$ satisfies $\mathcal{P}$, or $\Xi$ does not decompose as the join of a complete graph with copies of the disjoint union of two single vertices.

We begin by considering the case of right-angled Coxeter groups. Namely:

Proposition 1**.**

Let $\Gamma$ be a finite simplicial graph. Then $\mathrm{Aut}(C_{\Gamma})$ satisfies $\mathcal{P}$ if and only if $\Gamma$ does not decompose as the join of a complete graph with copies of the disjoint union of two single vertices.

The proposition will be essentially deduced from the following observation, which is a consequence of [6, Proposition 17.2.1]:

Lemma 1**.**

Let $\Gamma$ be a finite simplicial graph. The following assertions are equivalent:

• •

the right-angled Coxeter group $C_{\Gamma}$ is large (ie., it contains a finite-index subgroup which surjects onto a non-abelian free subgroup);

• •

$\Gamma$* does not decompose as the join of a complete graph with copies of the disjoint union of two single vertices;*

• •

$C_{\Gamma}$* decomposes as a direct sum of copies of $\mathbb{Z}_{2}$ and $\mathbb{D}_{\infty}$.*

Proof of Proposition 1..

A SIL in $\Gamma$ is the data of two vertices $u,v\in V(\Gamma)$ satisfying $d(u,v)\geq 2$ such that $\Gamma\backslash(\mathrm{link}(u)\cap\mathrm{link}(v))$ contains at least one connected component which does not contain $u$ nor $v$. Notice that, if $\Gamma$ decomposes as the join of a complete graph with copies of the disjoint union of two single vertices, then $\Gamma$ cannot contain a SIL.

If $\Gamma$ contains a SIL, then $\mathrm{Out}(C_{\Gamma})$ is large according to [24], ie., it contains a finite-index subgroup which surjects onto a non-abelian free subgroup. We deduce from the conditions (i), (ii) and (iii) that $\mathrm{Aut}(C_{\Gamma})$ has to satisfy $\mathcal{P}$.

From now on, assume that $\Gamma$ does not contain a SIL. As a consequence of [12, Theorem 1.4], $\mathrm{Out}(C_{\Gamma})$ must be finite, so that $\mathrm{Inn}(C_{\Gamma})$ must have finite index in $\mathrm{Aut}(C_{\Gamma})$. Notice that, if $\Lambda$ denotes the subgraph of $\Gamma$ generated by the vertices which are adjacent to all the vertices of $\Gamma$, then the center of $C_{\Gamma}$ is generated by the generators labelling the vertices of $\Lambda$. Therefore, $C_{\Gamma}$ decomposes as $C_{\Gamma\backslash\Lambda}\oplus\mathbb{Z}_{2}^{\#V(\Lambda)}$ and $\mathrm{Inn}(C_{\Gamma})\simeq C_{\Gamma}/Z(C_{\Gamma})\simeq C_{\Gamma\backslash\Lambda}$.

We distinguish two cases. If $\Gamma\backslash\Lambda$ does not decompose as the join of disjoint unions of two single vertices, then it follows from Lemma 1 and from the conditions (i), (ii) and (iii) that $\mathrm{Aut}(C_{\Gamma})$ has to satisfy $\mathcal{P}$. Otherwise, if $\Gamma\backslash\Lambda$ decomposes as the join of disjoint unions of two single vertices, then $C_{\Gamma\backslash\Lambda}$ is a direct sum of infinite dihedral groups. As such a group is virtually abelian, we deduce from the conditions (iii) and (v) that $\mathrm{Aut}(C_{\Gamma})$ cannot satisfy $\mathcal{P}$. ∎

We are now ready to prove Theorem F.

Proof of Theorem F.

If $\Gamma$ contains two non-adjacent vertices $u$ and $v$ which are not both labelled by $\mathbb{Z}_{2}$, then we have seen in the previous section that $\mathrm{PConjAut}(G_{u}\ast G_{v})$ is a quotient of $\mathrm{PConjAut}(\Gamma\mathcal{G})$ and that it acts on a simplicial tree with a WPD isometry. Because $G_{u}\ast G_{v}$ is not virtually cyclic, it follows from [22] that $\mathrm{PConjAut}(G_{u}\ast G_{v})$ must be acylindrically hyperbolic, so that $\mathrm{ConjAut}(\Gamma\mathcal{G})$ has to satisfy $\mathcal{P}$ as a consequence of the conditions (i), (ii) and (iii).

Now, assume that any two non-adjacent vertices of $\Gamma$ are labelled by $\mathbb{Z}_{2}$. As a consequence, if we decompose $\Gamma$ as a join $\Lambda\ast\Xi$ where $\Lambda$ is the subgraph generated by the vertices with are adjacent to all the vertices of $\Gamma$, then the vertices of $\Xi$ have to be all labelled by $\mathbb{Z}_{2}$. Therefore, $\mathrm{PConjAut}(\Gamma\mathcal{G})$ is isomorphic to the direct sum

[TABLE]

As a consequence of the condition (ii) and (iv), we know that $\mathrm{ConjAut}(\Gamma\mathcal{G})$ satisfies $\mathcal{P}$ if and only if so does one the direct factors above. Therefore, it follows from Proposition 1 that $\mathrm{ConjAut}(\Gamma\mathcal{G})$ does not satisfy $\mathcal{P}$ only when $G_{u}/Z(G_{u})$ does not satisfy $\mathcal{P}$ for every $u\in\Xi$ and when $\Xi$ decomposes as the join of a complete graph with copies of the disjoint union of two single vertices. ∎

Remark 2**.**

It is worth noticing that the proof above shows that Theorem F still holds if the condition (i) satisfied by $\mathcal{P}$ is replaced with: if $A$ and $B$ are two groups which are not both cyclic of order two, then $\mathrm{ConjAut}(A\ast B)$ satisfies $\mathcal{P}$.

Finally, Theorems B, C and D are direct consequences of the combination of Theorem E above with our next statement:

Proposition 2**.**

The following group properties satisfy the conditions (i), (ii), (iii), (iv) and (v) above:

• •

being SQ-universal;

• •

virtually having many quasimorphisms;

• •

not being boundedly generated.

Proof.

It is clear that being SQ-universal satisfies the conditions (ii) and (v). The conditions (i), (iii) and (iv) follow from [22, Theorem 8.1], [21] and [11, Lemma 1.12] respectively.

It is clear that not being boundedly generated satisfies the conditions (ii), (iii), (iv) and (v). About condition (i), we refer to the discussion following [4, Theorem 2.33] and to the references mentioned therein.

Virtually having many quasimorphisms satisfies the conditions (ii), (iii) and (iv), see [11, Proposition 1.18] for more details. Condition (i) is proved by [3]. About condition (v), it is sufficient to notice that $\mathbb{Z}$ has a one-dimensional space of quasimorphisms and next to apply condition (iv). ∎

We conclude the article by proving Proposition E.

Proof of Proposition E..

According to [9, Theorem 3.15], $\mathrm{ConjAut}(\Gamma\mathcal{G})$ has finite index in $\mathrm{Aut}(\Gamma\mathcal{G})$. Therefore, the equivalences between (i), (iii), (iv) and (v) are immediate consequences of Theorems B, C and D and Proposition 2.

Next, it follows from [11, Corollary 1.6] that involving all finite groups satisfies the conditions (ii), (iii) and (iv) above. Condition (v) is also clearly satisfied. Now let $A$ and $B$ be two finite groups which are not both cyclic of order two. Notice that $\mathrm{ConjAut}(A\ast B)$ contains $\mathrm{Inn}(A\ast B)\simeq A\ast B$ as a normal subgroup, which is virtually a finitely generated free group of rank $\geq 2$ according to [25, Proposition 4, page 6]. As a consequence of [11, Lemma 1.5], $\mathrm{ConjAut}(A\ast B)$ has to involve all finite groups. Therefore, we deduce from Remark 2 that the conclusion of Theorem F holds in the realm of graph products of finite groups, proving the equivalence between (i) and (ii) in our proposition.

If we denote by $\langle\Lambda\rangle$ (resp. $\langle\Xi\rangle$) the subgroup of $\Gamma\mathcal{G}$ generated by the vertex-groups labelling the vertices of $\Lambda$ (resp. of $\Xi$), then $\Gamma\mathcal{G}$ decomposes as the direct sum $\langle\Xi\rangle\oplus\langle\Lambda\rangle$. Notice that $\langle\Lambda\rangle$ is finite, so that $\langle\Xi\rangle$ must have finite index in $\Gamma\mathcal{G}$. If $\Xi$ contains a vertex $u$ which is labelled by a group $G_{u}$ which is not cyclic of order two, then $\Gamma\mathcal{G}$ contains the free product $\langle G_{u},G_{v}\rangle\simeq G_{u}\ast G_{v}$ as a subgroup, where $v$ is a vertex of $\Gamma$ which is not adjacent to $u$. But such a group is virtually a non-abelian free group according to [25, Proposition 4, page 6], so $\Gamma\mathcal{G}$ cannot be virtually abelian. So let us suppose that the vertices of $\Xi$ are all labelled by $\mathbb{Z}_{2}$. As a consequence, the subgroup $\langle\Xi\rangle$ is isomorphic to the right-angled Coxeter group $C_{\Xi}$, so that the desired conclusion follows from Lemma 1. ∎