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Abstract
A finite group is called Involutive YangBaxter (IYB) if there exists a bijective cocycle for some module . It is known that every IYBgroup is solvable, but it is still an open question whether the converse holds. A characterization of the IYB property by the existence of an ideal in the augmentation ideal complementing the set lead to some speculation that there might be a connection with the isomorphism problem for . In this paper we show that if is a nilpotent group of class two and is an IYBgroup of order coprime to that of , then is IYB. The class of groups that can be obtained in that way (and hence are IYB) contains in particular Hertweck’s famous counterexample to the isomorphism conjecture as well as all of its subgroups. We then investigate what an IYB structure on Hertweck’s counterexample looks like concretely.
On the IYBproperty in some solvable groups]On the IYBproperty in some solvable
groups
[
Mathematics Subject Classification (2010). 20C05, 16S34.
Keywords. Involutive YangBaxter Groups, Integral Group Rings.
Recently there has been considerable interest in the problem of characterizing those finite groups for which there exists a bijective cocycle ( being a finite module with ). A group for which such a cocycle exists is called an InvolutiveYangBaxter group. It is easily seen that preimages of submodules of are subgroups of , and hence Hall subgroups of all possible orders exist in (since submodules of corresponding order exist in ). By a wellknown theorem of Hall it thus follows that IYBgroups are solvable. This was first observed in [ESS99], which is also the article that first introduced the notion of an IYBgroup. Whether, conversely, every solvable group is IYB is an open question. There is an equivalent formulation of the IYB property, stating that a group is IYB if and only if there is a left ideal in the augmentation ideal such that the elements of the form form a complement of , that is, they form a set of residue class representatives for the quotient . The relation with bijective cocycles is evident: given such an ideal , the map which maps to is a bijective cocycle. Now if were a twosided ideal, then would be the circle group of the radical ring and hence it would be determined by its group ring. But even if is not twosided, the group complements the trivial units in the unit group . However it is not clear what implications, if any, the existence of a (nonnormal) complement to the trivial units has. In the introduction to [CJO12] it is conjectured that there is a connection between the question whether every solvable group is IYB and the integral isomorphism problem. In this short article we show that the counterexample to the isomorphism problem given by Hertweck in [Her01] cannot serve as an example for a solvable group which is not IYB (and we also exclude a plethora of similarly constructed groups). This is done in some generality in Theorem \@setrefthm_semidirect_nilpot, which states that the semidirect product of a nilpotent group of class two with any IYBgroup is again IYB. A deeper connection with the isomorphism problem therefore seems unlikely. We also give an explicit description of a bijective cocycle on Hertweck’s counterexample to the isomorphism problem. This seems interesting as it shows that in the context of IYBgroups there is apparently nothing special about Hertweck’s group, which reinforces the conjecture that all soluble groups might be IYB.
Definition 2.1.

A finite group is called Involutive YangBaxter (IYB) if there is a (left) module and a bijective cocycle . We call the pair an IYBstructure on the group .

Assume we are given, in addition to , a group which acts on (from the left) by automorphisms. If is a module and is a bijective cocycle with the property that
(\theequation) then we call the pair an equivariant IYBstructure on . Note that a equivariant IYBstructure simply an ordinary IYBstructure.

Let and be two equivariant IYBstructures on . Then and are called isomorphic if there is a module isomorphism such that .
The following two propositions show that a semidirect product with factors of coprime orders is IYB if and only if has an equivariant IYBstructure and is IYB. For a group that has an automorphism group with a large part there may be significantly fewer IYBstructures equivariant with respect to these automorphisms than there are IYBstructures in general. This is evidenced for instance by Theorem \@setrefthm_upper_triang below, as well as by abelian groups where the situation is similar. So finding a group which has no equivariant IYBstructure with respect to some solvable group acting on it by automorphisms would lead to a solvable group which is not IYB. However, at this point, it is unclear whether such a group exists, and if so, where to look for it.
Proposition 2.2 (see also [CJdR10, Theorem 3.4]).
Let be a finite group. If is IYB and has an equivariant IYBstructure, then is IYB.
Proof.
Let be an IYBstructure on , and let be an equivariant IYBstructure on . Note that is a module by definition, and can be construed as a module by letting act trivially. Then is a module and defines a bijective cocycle on . ∎
The following proposition shows that the converse of the above is true if is a normal Hallsubgroup of , that is, if .
Proposition 2.3.
Assume is IYB and assume that is a Hallsubgroup of . Then and are IYB, and has an equivariant IYBstructure.
Proof.
Let be an IYBstructure on . Decompose , where and (clearly this is possible since is an module with ). Since is IYB it must in particular be solvable, and therefore all Hallsubgroups of a given order are conjugate. Moreover the preimages of submodules of under form subgroups of (this is an elementary computation). Since is a normal Hallsubgroup it follows that . Moreover there is a such that . The map is also a bijective cocycle, and . Clearly the restricted maps and are bijective cocycles. All that is left to verify if that is equivariant. But
(\theequation) 
Since is a direct sum it follows . (As a side note: it also follows that acts trivially on ) ∎
In order to prove Theorem \@setrefthm_semidirect_nilpot below we need a few facts on the augmentation ideal in an integral group ring. In what follows we denote the augmentation ideal in a group ring by and its th power by .
Remark 2.4 (see [Ms02, Lemma 9.3.6 and Corollary 9.3.7] and [Pas68, Theorem 7.1]).
We are going to need the following two wellknown facts. Let be a finite group.

There is an isomorphism of abelian groups
(\theequation) 
and (the more general assertion that is equal to the th term in the lower central series is, or rather was, known as the dimension subgroup conjecture; it is wrong in general, but true for odd order groups).
Theorem 2.5 (see [San72, Theorem 1.24]).
Let be a finite group. Then the embedding
(\theequation) 
splits (as a homomorphism of abelian groups).
Theorem 2.6.
Assume is a nilpotent group of class two with a group of coprime order acting on it by automorphisms. Then possesses an equivariant IYBstructure. In particular, is IYB if and only if is IYB.
Proof.
We can assume without loss that is a group for some prime which does not divide the order of . Let denote the adic integers. Consider the module . We claim that the set (that is, the elements for ) form an submodule of . Clearly maps elements of the form to elements of the same form, so all we need to check is that these elements form a submodule (of course this was already implicitly used in the formulation of Theorem \@setrefthm_sandling). So assume . We get
(\theequation) 
Now we will show that is complemented in as a module. Since every module is relatively projective it follows that is complemented as a module if and only if it is complemented as a module. A more elementary way of stating this is that if is a projection to , then
(\theequation) 
is a projection onto as well, but . Therefore the kernel of is an module complement for . The existence of a module complement is precisely what is asserted in Theorem \@setrefthm_sandling. Since acts trivially on the complement for we just obtained is automatically a module. Hence we have obtained an stable ideal such that (here we use the second part of Remark \@setrefrem_dim_subgroup) and
(\theequation) 
where the first part of Remark \@setrefrem_dim_subgroup was used. It follows that is a complement for the set . Therefore the map
(\theequation) 
is a bijective cocycle which is equivariant due to being stable under the induced action of on . ∎
The following is another way of constructing an equivariant IYBstructure on nilpotent groups of class two, provided the order of the group is odd.
Remark 2.7 (see [Aw73] or [Cjo12, Proposition 9.4]).
For a group of odd order there is a wellknown IYBstructure on , by defining the following addition on
(\theequation) 
and letting act from the left by the formula (owed to the fact that the elements of , when construed as an module, should correspond to the elements in the corresponding quotient of the augmentation ideal of the group ring). It is clear by definition that this IYBstructure is equivariant. Of course this construction only works when is of odd order, because only then will square roots of group elements necessarily exist.
The following proposition is one of the ingredients required in order to construct an explicit IYBstructure on the counterexample to the isomorphism problem given by Hertweck. It is a slight generalization of [CJdR10, Corollary 3.5], the latter stating that given two groups and , both being IYB and being a permutation group, their wreath product will be IYB as well.
Proposition 2.8.
Let be a finite group and let be a group acting on by automorphisms. Assume has an equivariant IYBstructure. Then ( factors) has an equivariant IYBstructure (where denotes the symmetric group on points).
Proof.
Let be an equivariant IYBstructure on . Then acts on both and in the natural way, by letting (where and ) send (an element of or ) to . We define
(\theequation) 
Clearly this is a bijective 1cocycle, and equivariance is also clear, since
∎ 
It is of course trivial that if has an equivariant IYBstructure, and is a homomorphism from another group into , then has a equivariant structure, where acts on in the same way as . So it follows more generally that if a finite group is IYB and it acts on through automorphisms induced by elements of , then is IYB. We can also write down its IYBstructure explicitly provided we know (explicitly) a equivariant IYBstructure on and an IYBstructure on .
In [Her01], M. Hertweck gave an example of a finite group such that for some other finite group not isomorphic to . On the other hand it is wellknown that the circle group of a radical ring (a concept closely related to IYBgroups) has the property that implies (see for instance [MS02, Chaper 9.4]). Therefore there was some hope that Hertweck’s group might be an example of a solvable group which is not IYB. However this is not the case. Hertweck’s counterexample is a semidirect product , where is a group of nilpotency class two, and is a group which is a semidirect product of two abelian groups. By Theorem \@setrefthm_semidirect_nilpot it is already clear that is indeed IYB. Note that the group with is constructed in the same way as , implying that it is IYB as well. In this section we are going to explain how to explicitly construct an IYBstructure on . There clearly is no problem constructing an IYBstructure on . The group is a direct product of four cyclic groups of order , and eight times a group which is described below. acts separately on the direct factors involving cyclic groups and those involving , acting through on the latter. Due to Proposition \@setrefprop_wreath it hence suffices to find an IYBstructure on which is equivariant with respect to the right subgroup of .
Now let be a prime such that , and let
(\theequation) 
where is central and (note that this is isomorphic to the Sylow subgroup of ). Define the following automorphisms of :
(\theequation) 
where is a generator of the multiplicative group . Note that , (since the stabilize the subgroup , but does not). Moreover and . Hence it follows that is isomorphic to . In particular it has order . It has a faithful representation over , which can be obtained by considering the action of on :
(\theequation) 
Note moreover that since the group contains a Sylow subgroup of . To see this note that has order , and is odd. Therefore contains a Sylow subgroup of . The elements in which act trivial on must send to , to and to for some . It is easily seen that the order of such an automorphism is either or one. Hence the Sylow subgroups of have the same order as those of .
Now we would like to describe an equivariant IYBstructure on . Together with \@setrefprop_wreath and the known construction of an IYBstructure on a semidirect product of abelian groups this is enough to piece together an IYBstructure on the group given by Hertweck. We will indeed see that, up to isomorphism, has a unique equivariant IYB structure. While this uniqueness is not necessary as far as constructing an IYBstructure on Hertweck’s counterexample goes, it shows that there is not much room for different equivariant IYBstructures (whereas computational experiments seem to indicate that nonequivariant IYBstructures on groups exist in abundance).
Theorem 3.1.
Let be the vector space with the following left action of :
(\theequation) 
Then the map
(\theequation) 
defines an equivariant IYBstructure on , and is up to isomorphism the only equivariant IYBstructure on .
Proof.
We start with an arbitrary equivariant IYBstructure on and show that it has to be of the claimed form. First note that contains the element of order . The adic group ring decomposes as a direct sum of the unramified extension and a number of copies of . The latter correspond to nonfaithful representations. Hence any faithful module has a direct summand of the form with . Since acts faithfully on it must act in such a manner on as well, forcing to be the direct sum (as all other possibilities would have a cardinality that is too large). That is, has to be an vector space.
Next note that the kernel of the action of on must be a normal subgroup of which is stable under the action of . There are only three such subgroups of , namely , and itself. Clearly must act nontrivially on , since otherwise would be an isomorphism between and the additive group of . If were to act faithfully on , then it would map onto a Sylow subgroup of . Then would be embedded in the normalizer of such a Sylow subgroup. This is impossible, since such a normalizer is isomorphic to the group of invertible upper triangular matrices over , and there is no element of order in that group.
We have hence established that is a faithful module. Clearly cannot be semisimple (because then would act trivially). Assume . Then would be a flag which must be stabilized by , again forcing into a subgroup of isomorphic to the group of invertible upper triangular matrices, which was impossible. Hence and is either one or two dimensional and semisimple. Assume is twodimensional. Then (since otherwise would be semisimple). But for any and any we have , which implies . But then for all , which implies that is a onedimensional subspace of . But must fix this subspace (since fixes ), and therefore fixes the flag , which is impossible by the same arguments as before. It follows hence that is onedimensional. The dual of is thus an module with simple top and radical length two. This means that this dual is an epimorphic image of , and it is in fact isomorphic since the latter also has dimension . Hence is isomorphic (as an module) to the dual of . So we can assume without loss that and act on as claimed. Write for the action of on . Then it follows, using the identity for all and , that an element acts in the following way
(\theequation) 
where is a group homomorphism and are arbitrary functions from to . We can conjugate the matrices in such a way that and both become the zero map (just find an stable complement for the subspace generated by the first standard basis vector).
Since we already know that maps into the socle of we can say that must be a nonzero multiple of the first standard basis vector. By applying an automorphism of we can hence choose to be the first standard basis vector, and therefore for all . The equivariance of now implies . Now we determine the images and . The fact that fixes implies that lies in the eigenspace of with associated eigenvalue one. So for some . In the same manner one sees that for some . Using we can conclude (again by equivariance). By comparing and we infer that , i. e. . Formula (\@setrefeqn_sdjkdjlkjd) now follows easily: . ∎
There are actually not that many groups of small order which cannot be seen to be IYB using Theorem \@setrefthm_semidirect_nilpot. Using the SmallGroups library that comes with the computer algebra system Gap ([GAP13]) one can show that the potential counterexamples of order have orders , , , and (excluding prime powers). For all except the groups of order one can compute an IYBstructure using a bruteforce approach (a slightly more intelligent approach might actually work for the groups of order 192 as well). For (small) groups one can use a heuristic approach to compute an IYBstructure. Namely, given a group one can pick a normal subgroup of order . Assume we already computed an ideal complementing . Then we can take the preimage of under the natural epimorphism and try to find a maximal submodule in this preimage which does not contain . If such a maximal submodule exists then it is indeed an ideal in complementing . While such a maximal submodule does not always exist, trying this multiple times with different subgroups and different choices of will typically work (actually no potential counterexamples were found in this way). Using an implementation of this heuristic in Gap it was possible to show that all groups of order up to (and including) are IYB. Also all other groups of order strictly less than turned out to be IYB. So the evidence that all nilpotent or even all solvable are IYB seems to be piling up.
This research was supported by the Fonds Wetenschappelijk Onderzoek  Vlaanderen (FWO) project G.0157.12. I would also like to thank F. Cedó for looking at a preliminary version of this paper and suggesting Remark \@setrefrem_cedo.
References
 [AW73] J. C. Ault and J. F. Watters. Circle groups of nilpotent rings. Amer. Math. Monthly, 80:48–52, 1973.
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