Right-angled Artin groups as finite-index subgroups of their outer automorphism groups

We prove that every right-angled Artin group occurs as a finite-index subgroup of the outer automorphism group of another right-angled Artin group. We furthermore show that the latter group can be chosen in such a way that the quotient is isomorphic to ( ℤ ∕2 ℤ ) 𝑁 for some 𝑁 . For these, we give explicit constructions using the group of pure symmetric outer automorphisms. Moreover, we need two conditions by Day–Wade and Wade–Brück about when this group is a right-angled Artin group and when it has finite index.

notation Γ −  for the induced subgraph with vertex set (Γ) ⧵ .When we talk of a component of a graph, we always mean a connected component.Furthermore, we use lk() for the set of neighbours of  and st() for the union of lk() with  itself.

2.2
Right-angled Artin groups Definition 2.1.For a given non-empty graph Γ, the right-angled Artin group  Γ has the following presentation: Recall that [, ] ∶=  −1  −1 is a notation for the commutator of  and .So, the generators of  Γ correspond to the vertices of Γ and two generators commute if there is an edge between the two corresponding vertices in the graph Γ.
Next, we give a set of generators for the automorphism group of a right-angled Artin group  Γ .This is based on [14, subsection 2.5].We refer the reader to this source for more details.For a proof of the statement, one can look at [13] and [12], which are the original sources of this theorem.Servatius conjectured it and proved it for some special cases and Laurence gave a proof for general graphs.Note that the terminology in [13] and [12] is slightly different to the one we use here.
For simplicity, we assume that the graph Γ consists of vertices  1 , … ,   .There are four types of automorphisms that together generate Aut( Γ ).
• Γ -legal transvections: For vertices   and   that satisfy the condition lk(  ) ⊆ st(  ), we define the automorphism   , of  Γ by mapping the generators as follows:   ↦     and   ↦   for  ≠ .These are called Γ-legal (left) transvections.Analogously, again, for   and   with lk(  ) ⊆ st(  ), one can define the Γ-legal right transvection   , by mapping   ↦     and   ↦   for  ≠ .
• Partial conjugations: Another type of generators are the so-called (Γ-legal) partial conjugations.
Here, we often omit the term Γ-legal as we do not introduce other partial conjugations.For a vertex   ∈ (Γ) and a component  of Γ − st(  ), we define the partial conjugation    by   ↦      −1  for   ∈  and   ↦   for   ∉ .• Γ -legal permutations: In order to get another automorphism of  Γ , we can permute the generators.However, not all permutations give automorphisms, but only those that correspond to automorphisms of the graph Γ.We call these Γ-legal permutations.Due to the fact that these automorphisms of  Γ are related to graph automorphisms of Γ, Γ-legal permutations are often also called 'graph automorphisms'.• Inversions: For  ∈ {1, … , }, we define the inversion   as follows:   ↦  −1  and   ↦   for  ≠ .
Remark.In a general setting, we write     ,  instead of   , .The same applies to the notation for the other types of generators.
We use the notation from above also for the images of the generators in Out( Γ ).These elements then generate Out( Γ ).

Pure symmetric outer automorphisms
The following is based on [8,Section 2.2].An automorphism of  Γ is called pure symmetric if every generator   is mapped to a conjugate of itself.Note that the conjugating element may depend on   .The set of such elements forms a subgroup of Aut( Γ ) as the composition of two pure symmetric automorphisms is again pure symmetric.This subgroup is called the group of pure symmetric automorphisms and denoted by PSA( Γ ).A generating set for it is the set of partial conjugations; see [8,Theorem 2.5].We define PSO( Γ ), the group of pure symmetric outer automorphisms, as the image of PSA( Γ ) in Out( Γ ).
Our goal is to first describe conditions for when PSO( Γ ) has finite-index in Out( Γ ) and for when it is a right-angled Artin group.Then, we show in Section 3 that every right-angled Artin group  Λ occurs as PSO( Γ ) for some graph Γ for which PSO( Γ ) has finite index in Out( Γ ).

2.3.1
Finite-index condition for PSO( Γ ) The following theorem is based on [15, Appendix A].
Theorem 2.2.Let Γ be a graph.Then the condition is equivalent to the group PSO( Γ ) having finite index in Out( Γ ).
Note that condition (1) is equivalent to the group  Γ having no Γ-legal transvections.The idea for the proof that condition (1) implies finite index is to show that where Inv and Per denote the subgroups of Out( Γ ) generated by the inversions, respectively, by the Γ-legal permutations.For the other direction, one can show that the existence of a Γ-legal transvection implies that the quotient has infinitely many elements.

2.3.2
Condition for when PSO( Γ ) is a right-angled Artin group In this subsection, we state a condition about when PSO( Γ ) is a right-angled Artin group.This condition was developed and proved by Day and Wade in [8].This is also the source for this subsection, in particular, [8, Chapters 2 and 5].We first need the following two definitions.Remark.Note that the definition above is not symmetric.That is, for two components  and  of Γ − st(), the fact that there is a  ∈  such that  is also a component of Γ − st() is not equivalent to the fact that there is an  ∈  such that  is also a component of Γ − st().There is an edge between  and  in the support graph if any of these two conditions holds.
Remark.As in Definition 2.4, we often use the same symbol for the vertices in the support graph  Γ  and the components of Γ − st(), even though these are not the same.
For  ∈ Γ − st(), we often write the component of Γ − st() that contains  as []  .So, equivalently, there is an edge between two vertices  and  in the support graph  Γ  if there is a  ∈ Γ − st() such that  = []  and  is also a component of Γ − st().
We now define the following graph Θ (depending on Γ).This graph has two types of vertices.We call them vertices of types I and II.More precisely, for every vertex  ∈ (Γ), we have the following vertices.
• Vertices of type I: For every edge  in  Γ  , we have a vertex    .• Vertices of type II: We also have a vertex for every component of  Γ  except one, that is, we have vertices   1 , … ,   min(()−1,0) , where () is the number of components of  Γ  .We take the minimum with 0 to avoid the special case () = 0.This happens if Γ − st() has no vertices and thus also We now state the theorem about when PSO( Γ ) is a right-angled Artin group.
Theorem 2.5 [8,Theorem 5.12].Let Γ be a graph.Then PSO( Γ ) is a right-angled Artin group if and only if all support graphs of Γ are forests.In this case, PSO( Γ ) ≅  Θ for Θ as defined above.
Remark.Note that in this theorem, in contrast to Definition 2.1, we also treat the group {id} =  ∅ as a right-angled Artin group, where ∅ denotes the graph with no vertices.Namely, Θ is the empty graph if all support graphs have at most one vertex.This happens, for example, when Γ is a complete graph.Note that it makes sense that PSO( Γ ) is the trivial group in this case since when Γ is complete, then  Γ is ℤ |(Γ)| .Thus, conjugation by any element does nothing, so there are no pure symmetric outer automorphisms except the identity.
In [8,Chapter 5], one can find an explicit construction of the isomorphism between the rightangled Artin group  Θ and the group PSO( Γ ) in case all support graphs of Γ are forests.They define the vertices of type II more precisely in [8,Definition 5.4], describe the needed generators of PSO( Γ ) in [8, Section 5.1] and give the correspondence between the generators of  Θ and the generators of PSO( Γ ) in [8, Proposition 5.5].

RIGHT-ANGLED ARTIN GROUPS AS FINITE-INDEX SUBGROUPS OF 𝐎𝐮𝐭(𝑨 𝚪 )
In this section, we use Theorems 2.2 and 2.5 to show that every right-angled Artin group is a finiteindex subgroup of the outer automorphism group of some other right-angled Artin group.This is Theorem A, which we here restate as Theorem 3.1.Theorem 3.1.For any graph Λ, there is a graph Γ = Γ(Λ) such that 1.  Λ ≅ PSO( Γ ) and 2. PSO( Γ ) has finite index in Out( Γ ).
Remark.We first want to comment on how this theorem and also Theorem 3.4 below were developed.We used computer programs that can be found on [16].Using these, we found out that for any graph with at most four vertices, the corresponding right-angled Artin group occurs as PSO( Γ ) for some graph Γ.We then generalised these examples step by step until we arrived at the constructions given below.These computer programs are based on and contain parts of programs written by Benjamin Brück, which can be found on [5] and were used for [15].
We assume without loss of generality that (Λ) = { 1 , … ,   }.We also assume  ⩾ 3. The other cases are covered in the Appendix.For a given graph Λ, we define the graph Γ = Γ(Λ) as follows.We obtain Γ from Λ by adding vertices  1 ,  2 ,  1 ,  2 ,  1 , … ,   ,  1 and  2 and the edges depicted in Figure 1: The vertex  1 is connected to  1 , to all vertices   and to the vertex  1 , but not to the vertices   for  > 1.The vertex  2 is connected to  2 , all   and all   except  1 .The vertices   are connected to  1 ,  2 and the vertices   and  +1 , where  +1 ∶=  1 .In addition to the already defined edges, the vertices   are connected to  1 and  2 .Furthermore, we have the edges { 1 ,  1 }, { 2 ,  2 } and { 1 ,  2 }.Finally, if there are edges in Λ, these are also present in Γ.But as we work with an arbitrary graph Λ, we did not draw them in Figure 1.
We want to prove the following two lemmas, which together imply Theorem 3.1.Proof of Lemma 3.2.We want to use Theorem 2.5.So, we first need to show that every support graph is a forest.To compute the support graphs, we need to consider Γ − st() for all vertices  of Γ.Some of these are shown in Figure 2. The vertex  is coloured in red, the rest of st() is coloured in light red and all other deleted edges are coloured in grey.Note that for Figure 2c, it might be that also some of the vertices { 2 , … ,   } should be coloured light red, namely all vertices in st Λ ( 1 ).For the vertices that we did not consider in Figure 2, Γ − st() is analogous to one of the vertices we considered there.We can see that for all vertices  that are not in Λ, Γ − st() is connected, and thus, the support graph  Γ  consists of one vertex.Here, we used that  ⩾ 3, otherwise Γ − st( 1 ), respectively, Γ − st( 2 ) have two components as  2 , respectively,  1 become isolated.For the vertices   , the graph Γ − st(  ) has two components, one of which is { 1 ,  2 }.The second component contains all other vertices that are not in the star of   .Using the definition of the support graph, this means that  Γ   has two vertices.Whether there is an edge between them depends on whether (Λ) ⧵ st Λ (  ) is empty or not.If (Λ) ⧵ st Λ (  ) ≠ ∅, then there is a   ∈ (Λ) ⧵ st Λ (  ) and { 1 ,  2 } is also a component of Γ − st( ).Thus, by Definition 2.4, in this case, there is an edge in the support graph.However, if (Λ) ⧵ st Λ (  ) = ∅, we cannot satisfy the condition of Definition 2.4, so the support graph has no edge.
In summary, we see that all support graphs are forests, and thus, PSO( Γ ) is a right-angled Artin group by Theorem 2.5.As in this theorem, we call the underlying graph for this right-angled Artin group Θ.Furthermore, we know from the structure of the support graphs that there is exactly one vertex   in the graph Θ for every vertex   in the graph Γ. Depending on whether (Λ) ⧵ st Λ (  ) = ∅ or not, it is a vertex of type II or I.If (Λ) ⧵ st Λ (  ) = ∅, then where   is the edge in the support graph  Γ   .We claim that we have   ∼   if and only if   ∼   , which shows that Λ is isomorphic to Θ.In order to show this claim, we distinguish whether   is adjacent to   or not.We now prove the second lemma, which states that for this Γ = Γ(Λ), we have that PSO( Γ ) has finite index in Out( Γ ).
Proof of Lemma 3.3.In order to prove this, we want to use Theorem 2.2, so we need to show that lk() ⊆ st() implies  = .
We do this with Table 1, which lists all pairs (lk(), st()).If lk() ⊆ st(), we write a '✓'; otherwise, we provide an element in lk() ⧵ st().Regarding the indices,  and  go from 1 to  and  and  go from 2 to .For the parts denoted with ' * ', we need to distinguish what the indices are, which we do in the following.All these special cases need the assumption  ⩾ 3. The problem with  = 2 is that we only have two vertices   and both of them are connected to all vertices of Λ.We also need this assumption in the table, namely that  2 ≠   and thus,  2 ∉ st( 1 ).
With the table and the special cases, we conclude that lk() ⊆ st() implies  = .As discussed in the beginning, this concludes the proof.□ Next, we provide a better construction to strengthen Theorem 3.1.
Proof sketch of Lemma 3.7.To show this, one uses the fact that a graph automorphism sends adjacent vertices to adjacent vertices and, in particular, preserves the degree of every vertex.One can first show that the levels of Γ ′ are fixed, that is, the vertices   are mapped to vertices   by any automorphism, and so on.This is done from bottom to top, that is, starting with the   , then the   , and so on.Next, one can show that this implies that the vertices also need be fixed pointwise by any automorphism, which concludes the proof.□ We finish this section with a corollary about the structure of the quotient Out( Γ ′ )∕ PSO( Γ ′ ) and compute the index of PSO( Γ ′ ) in Out( Γ ′ ).We still assume  ⩾ 3.One can prove a similar version of this corollary for  < 3 using the constructions in the Appendix.Corollary 3.8.For the graph Γ ′ = Γ ′ (Λ) as described before, and thus, the index of PSO( Γ ′ ) in Out( Γ ′ ) is 2  , where  = 2 + 9 is the number of vertices of Γ ′ .
Proof.Using the generators of Aut( Γ ′ ) from Section 2.2, one can show that if PSO( Γ ′ ) has finite index in Out( Γ ′ ), one has Here, Inv is the subgroup of Out( Γ ′ ) generated by inversions and Per is the subgroup generated by Γ ′ -legal permutations.Since Γ ′ has no non-trivial graph automorphisms, the group Per of Γ ′legal permutations contains only the identity.Using that Inv ≅ (ℤ∕2ℤ)  for  the number of vertices of Γ ′ , we get that As |(ℤ∕2ℤ)  | = 2  , we also get that the index of PSO( Γ ′ ) in Out( Γ ′ ) is 2  .To complete the proof, it remains to argue why  satisfies the claimed equality.The graph Γ ′ as constructed above has 2 + 9 vertices because Λ has  vertices and we add  + 9 vertices.□

CONCLUSION AND OUTLOOK
We conclude this paper by summarising what we achieved and pointing out some possible directions for further research.In the paper [8], Day and Wade gave a condition for when PSO( Γ ) is a right-angled Artin group and which one it is.In this paper, we asked the question which right-angled Artin groups occur as (finite-index) subgroups of Out( Γ ) when we vary Γ.Using the condition by Day-Wade and another condition by Wade-Brück, we answered this question in Theorem 3.1 by showing that every right-angled Artin group occurs as PSO( Γ ) for some graph Γ for which PSO( Γ ) has finite index in Out( Γ ).
A possible direction for further research is to simplify the construction or to impose further conditions as we did in Theorem 3.4, where we additionally wanted that the graph Γ has no non-trivial graph automorphisms.The advantage of this compared to the theorem above is that it simplifies the structure of the quotient Out( Γ )∕ PSO( Γ ).For other simplifications, one could try to reduce the number of vertices or to reduce the index of PSO( Γ ) in Out( Γ ).
Another interesting topic to think about is to go beyond the subgroup of pure symmetric outer automorphisms and investigate for which graphs Γ a fixed right-angled Artin group  Λ is a (finiteindex) subgroup of the outer automorphism group of  Γ .As mentioned above, we showed in this paper, using the subgroup of pure symmetric outer automorphisms, that every right-angled Artin group  Λ occurs as a finite-index subgroup of Out( Γ ) for some graph Γ.Also, for fixed Λ, one can, using the condition of Day-Wade, answer the question when PSO( Γ ) is isomorphic to  Λ .Namely,  Λ ≅ PSO( Γ ) if and only if the graph Γ satisfies that all its support graphs are forests and the graph described in Theorem 2.5 is isomorphic to Λ.It would be interesting to generalise this to other examples that do not depend on the subgroup of pure symmetric outer automorphisms and maybe for fixed Λ give conditions that a graph Γ needs to satisfy so that  Λ is a finite-index subgroup of Out( Γ ).Such results, even just for some graphs Λ, might give insights about how right-angled Artin groups interact with their outer automorphism groups and help to find more general results.However, a difficulty with this approach is that one cannot look at a fixed subgroup of Out( Γ ) but needs to consider all subgroups of Out( Γ ) at the same time.
Another interesting path one could take is to change the point of view and return to the question that Day and Wade asked; see [8,Question 1.1].This is to not ask for fixed Λ when  Λ is a (finiteindex) subgroup of the outer automorphism group of another right-angled Artin group, but to ask for fixed Γ which right-angled Artin groups occur as a (finite-index) subgroup of Out( Γ ) and if there are any at all.When restricted to the subgroup of pure symmetric outer automorphisms, this is again answered by the condition of Day-Wade, but it would be interesting to generalise this to arbitrary (finite-index) subgroups.
Finally, one could also try to find examples when the automorphism group Aut( Γ ) has rightangled Artin groups as (finite-index) subgroups.One could ask the same questions as the ones asked by Day-Wade and in this paper.There is already a result known in this direction.Namely, Charney-Ruane-Stambaugh-Vijayan prove in [7,Theorem 3.6] that the group PSA( Γ ) is a rightangled Artin group if the graph Γ has no separating intersection of links.For PSA( Γ ), one can still use the condition for finite index as in Theorem 2.2 because PSO( Γ ) has finite index in Out( Γ ) if and only if PSA( Γ ) has finite index in Aut( Γ ).To show this, one needs that Inn( Γ ) is a normal subgroup of Aut( Γ ) and that Inn( Γ ) ⊆ PSA( Γ ).Then, the natural map is well defined and bijective.Thus, since Out( Γ )∕ PSO( Γ ) = (Aut( Γ )∕ Inn( Γ ))∕(PSA( Γ )∕ Inn( Γ )), the index of PSO( Γ ) in Out( Γ ) is the same as the index of PSA( Γ ) in Aut( Γ ).One could try to use [7] and the finite-index condition from Theorem 2.2 to answer the question which rightangled Artin groups occur as a (finite-index) subgroup of Aut( Γ ) for some Γ.Similarly to this paper, one could use computer programs to find examples and then try to generalise these.

APPENDIX: SMALL GRAPHS
In this appendix, we give examples for the graphs with less than three vertices to complete the proofs of Theorems 3.1 and 3.4.For every such graph Λ, we give a graph Γ such that PSO( Γ ) ≅  Λ , PSO( Γ ) has finite index in Out( Γ ) and Γ has no non-trivial graph automorphisms.These graphs were found using computer programs that can be found on [16].This code can also be used to check that they satisfy the conditions of Theorems 3.1 and 3.4.In this appendix, we only sketch why the graph Θ as defined in Theorem 2.5 is indeed Λ.
Define Λ 1 as the graph with one vertex, Λ 2 as the graph with two vertices and no edge and Λ 3 as the graph with two vertices and an edge.Then, one can show that the graphs Γ 1 , Γ 2 and Γ 3 as in Figures A1a, A1b and A1c satisfy the conditions stated above, that is, PSO( Γ  ) ≅  Λ  (using Theorem 2.5), PSO( Γ  ) has finite index in Out( Γ  ) (using Theorem 2.2) and Γ  has no non-trivial graph automorphisms.
As mentioned above, we only sketch PSO( Γ  ) ≅  Λ  ; for the complete arguments, the computer programs from [16] can be used.For Γ 1 ,  9 is the only vertex with a separating star.The in the graph Θ 1 as defined in Theorem 2.5.For all other vertices   , Γ 1 − st(  ) is connected, and hence, they do not define vertices in Θ 1 .Thus, we have Θ 1 ≅ Λ 1 .For Γ 2 , the vertices  6 and  7 are the only ones with separating stars.Both their support graphs have two vertices and an edge, so the graph Θ 2 has two vertices   6  6 and   7  7 , where  6 and  7 are the edges in the support graphs of  6 and  7 , respectively.These two vertices satisfy the condition for when there is no edge between vertices of type I, and thus, the graph Θ 2 has no edge.Thus, we have Θ 2 ≅ Λ 2 .For Γ 3 , the vertices  8 and  9 are the only vertices with separating stars and their support graphs have both two vertices and no edge.Thus, in Θ 3 , we have two vertices  1 and an edge.Hence, we have Θ 3 ≅ Λ 3 .Using Theorem 2.5, we get that PSO( Γ  ) ≅  Λ  .

J O U R N A L I N F O R M AT I O N
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R E F E R E N C E S
Γ  has no vertices.Concerning the edges, vertices of type II are connected to all other vertices and vertices    and    of type I are connected except when (, ) is an SIL-pair and the edges are of the form  = {[]  , } and  = {[]  , }, where  is a component of both Γ − st() and Γ − st().

F I G U R E A 1
Proving Theorems 3.1 and 3.4 for the small graphs.support graph consists of two vertices without an edge, and thus, we have only one vertex   9 1 1. I. Agol, The virtual Haken conjecture, Doc.Math.18 (2013), 1045-1087.With an appendix by Agol, Daniel Groves, and Jason Manning.ISSN: 1431-0635.

Definition 2.4. Let
Γ be a graph and  ∈ (Γ) be a vertex of Γ.We define the support graph  Γ  as follows.For every component  of Γ − st(), there is a vertex in  Γ  .Two vertices  and  in  Γ  are connected by an edge if there is a vertex  ∈  such that  is also a component of Γ − st().
then   ∼   as   , respectively,   is of type II and thus connected to all other vertices.Otherwise, both are of type I, but (  ,   ) is not an SIL-pair as they are connected.Thus, we have   ∼   as well.  ≁   : Here, both (Λ) ⧵ st Λ (  ) and (Λ) ⧵ st Λ (  ) are non-empty.Thus, both   and   are of type I.More precisely,   ) ∩ lk(  )) has { 1 ,  2 } as a component because both  1 and  2 are in lk(  ) ∩ lk(  ).As   ≁   , we get that (  ,   ) is an SIL-pair.Furthermore, the edges in the support graphs of   and   are of the form ([  ]   , { 1 ,  2 }) and ([  ]   , { 1 ,  2 }).Hence, there is no edge between   and   .
Thus, we indeed have that Λ ≅ Θ.We can conclude that Λ ≅  Θ ≅ PSO( Γ ),which is what we wanted to show.□ TA B L E 1 Determining when lk() is a subset of st().