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On Symmetric Elements and SymmetricUnits in Group RingsEric Jespers a & Manuel Ruiz Marín ba Department of Mathematics , Vrije Universiteit Brussel , Brussel,Belgiumb Departamento Métodos Cuantitativos , e Informáticos, UniversidadPolitécnica de Cartagena , Cartagena, SpainPublished online: 03 Sep 2006.

To cite this article: Eric Jespers & Manuel Ruiz Marín (2006) On Symmetric Elements and SymmetricUnits in Group Rings, Communications in Algebra, 34:2, 727-736, DOI: 10.1080/00927870500388018

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Communications in Algebra®, 34: 727–736, 2006Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927870500388018

ON SYMMETRIC ELEMENTS AND SYMMETRIC UNITS INGROUP RINGS

Eric JespersDepartment of Mathematics, Vrije Universiteit Brussel, Brussel, Belgium

Manuel Ruiz MarínDepartamento Métodos Cuantitativos, e Informáticos, Universidad Politécnicade Cartagena, Cartagena, Spain

Let R be a commutative ring, G a group, and RG its group ring. Let � � RG → RG

denote the R-linear extension of an involution � defined on G. An element x in RG issaid to be symmetric if ��x� = x. A characterization is given of when the symmetricelements �RG�� of RG form a ring. For many domains R it is also shown that �RG��

is a ring if and only if the symmetric units form a group. The results obtained extendearlier work of Bovdi (2001), Bovdi et al. (1996), Bovdi and Parmenter (1997),Broche Cristo (2003, to appear), Giambruno and Sehgal (1993), and Lee (1999), whodealt with the case that � is the involution ∗ mapping g ∈ G onto g−1.

Key Words: Group ring; Involution; Symmetric elements.

2000 AMS Subject Classification: 16U60; 16W10; 16S34; 20C07.

1. INTRODUCTION

Let RG denote the group ring of a group G over a commutative ring R withidentity. Any group involution � � G → G can be extended R-linearly to a ringinvolution � � RG → RG. The fixed elements of � are often called the �-symmetricelements (or simply, symmetric elements when � is clear from the context). Theset of fixed elements of RG (respectively G) we denote by �RG�� (respectively G�).In general, �RG�� is not a subring of RG. However, it is a subring if and only if itis a commutative subring.

A natural involution on G is the classical involution ∗ which maps g ∈ G ontog−1. In case R is a field of characteristic not two and RG is semiprime, then it followsfrom a result of Giambruno and Sehgal (1993) that �RG�∗ is Lie nilpotent if andonly if it is commutative. Giambruno and Sehgal (1993) and Lee (1999) determinedwhen �RG�∗ is Lie nilpotent for arbitrary groups G and fields R of characteristicnot two. Very recently, Broche Cristo (2003, to appear) characterized in an elegant

Received October 15, 2004; Revised March 15, 2005. Communicated by M. Ferrero.Address correspondence to Manuel Ruiz Marín, Departamento Métodos Cuantitativos e

Informáticos, Universidad Politécnica de Cartagena, Paseo Alfonso XIII, 50, Facultad cc. Empressa,30203 Cartagena, Spain; Fax: +34-968325745; E-mail: manuel.ruiz@upct.es

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manner when for an arbitrary commutative ring R and group G the symmetricelements �RG�∗ form a commutative ring.

The group of units of the group ring RG we denote by ��RG� and thesymmetric units for an involution � on G we denote by ���RG�. Broche Cristo(2003, to appear) showed that if G is a periodic group and R is a commutativering of odd prime characteristic, then �RG�∗ is a commutative ring if and only if�∗�RG� is a group (and hence a commutative group). The same result holds if R isa field of prime characteristic p and G is a locally finite p-group (Bovdi et al., 1996),or if R is a G-favorable domain (e.g., an infinite field) and G is a periodic group(Bovdi, 2001).

In this article, we deal with the question when �RG�� is a commutative ringfor an arbitrary involution �. We obtain a complete characterization, and we alsoshow that there is a strong link between the property that the symmetric elementsform a ring and the property that symmetric units form a group.

Clearly, as an R-module, �RG�� is generated by

S = �g + ��g� � g ∈ G\G�� ∪G��

Therefore, �RG�� is a ring if and only if the elements in S commute. The center ofG is denoted by Z�G�, the additive commutator � − �, with � ∈ RG, is denoted�� �, and the multiplicative commutator ghg−1h−1 of g h ∈ G is denoted by �g h�.

Lemma 1.1. If �RG�� is a commutative ring, then G� ⊆ Z�G�. In particular,g��g� = ��g�g ∈ Z�G� for all g ∈ G.

Proof. Let g ∈ G� and h ∈ G. We have to prove that gh = hg. Since �RG�� iscommutative this is clear if h ∈ G�. So assume h �∈ G�. Then, again since �RG��is commutative,

0 = �g h+ ��h�� = gh+ g��h�− hg − ��h�g�

Hence gh = hg + ��h�g − g��h�. Clearly, gh �= g��h� as h �= ��h�. So gh = hg orgh = ��h�g. In the latter case, gh = ��h���g� = ��gh�. Hence, gh ∈ G� and thus gand gh commute. Consequently, �gh�g = g�gh� and thus gh = hg. �

Lemma 1.2. If �RG�� is a commutative ring, then for any two noncommuting elementsg, h ∈ G, one of the following properties holds:

1. gh = h��g� or gh = ��h�g,2. char�R� = 2, gh = ��g���h�, g2 h2 ∈ G�.

Proof. Let g h ∈ G and assume that gh �= hg. As �RG�� is commutative,Lemma 1.1 implies that g �∈ G� and h �∈ G�, and also

0 = �g + ��g� h+ ��h��

= gh+ g��h�+ ��g�h+ ��g���h�− hg − h��g�− ��h�g − ��h���g��

Since g h �∈ G�, we have that gh �= g��h� and gh �= ��g�h. Therefore, if gh �=��g���h� or if char�R� �= 2, then gh coincides with one of the elements h��g�, ��h�h,

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SYMMETRIC ELEMENTS IN GROUP RINGS 729

or ��h���g�. However gh = ��h���g� is impossible. Indeed, for otherwise, gh ∈ G�

and thus, by Lemma 1.1, gh ∈ Z�G� and thus gh = hg. This finishes the proof ofpart one.

So assume now that gh = ��g���h� (and thus ��h���g� = hg) andchar�R� = 2. Then

0 = �g + ��g� h+ ��h�� = g��h�+ ��g�h+ h��g�+ ��h�g�

Hence g��h� coincides with one of the elements in ���h�g h��g� ��g�h�. Ifg��h� = ��h�g, then also h��g� = ��g�h. Because of Lemma 1.1, g��g� ∈ Z�G�,and thus we obtain gh��g� = g��g�h = hg��g�. Hence gh = hg, a contradiction.If g��h� = h��g�, then g��h� = ��g��h�� and thus it is central. It follows that��h�g��h� = g��h���h� = h��g���h� and thus ��h�g = h��g� = g��h�. Howeverthis is excluded in the previous case. So we obtain that g��h� = ��g�h. Since g��g� =��g�g and because gh = ��g���h�, we obtain that g2��h� = g��g�h = ��g�gh =��g���g���h�. Therefore, g2 = ��g2� and hence g2 ∈ G�. Similarly, by multiplyingthe equality g��h� = ��g�h on the right by h, we obtain that h2 ∈ G�. This finishesthe proof. �

2. CHARACTERISTIC NOT 2

Throughout this section, R is a commutative ring with identity, andchar�R� �= 2. Further, � is an involution on a group G. From Lemma 1.2, we knowthat if �RG�� is a ring and g h ∈ G are noncommuting elements, then gh = h��g�or gh = ��h�g. We will show that then also gh = h��g� = ��h�g. But first, we needthe following lemma.

Lemma 2.1. Let g h ∈ G. Assume gh �= hg and h−1gh = ��g�. If �RG�� iscommutative and char�R� �= 2, then gh = ��h�g and g2 h2 ∈ G�.

Proof. Lemma 1.2, applied to the noncommuting elements gh and h, yields thatgh2 = h��h���g� or gh2 = ��h�gh.

We claim that in both cases gh = ��g���h�, and thus also hg = ��h���g�.Indeed, if gh2 = h��h���g� then, because h��h� = ��h�h ∈ Z�G�, we obtain thatgh2 = ��g���h�h. Hence, gh = ��g���h�. If gh2 = ��h�gh, that is, gh = ��h�g then,because of the hypothesis gh = h��g�, we obtain that gh = ��h�g = h��g�. Hence,gh2 = ��h�gh = ��h�h��g� = ��g���h�h and thus gh = ��g���h�.

So we have that gh = ��g���h�. Hence, the hypothesis h−1gh = ��g� impliesthat gh = h−1gh��h� = h−1h��h�g = ��h�g. Consequently,

��h�g = gh = h��g��

Multiplying by g on the right-hand side we get that ��h�g2 = h��g�g = hg��g� =��h���g2�. Thus g2 = ��g2� and hence g2 ∈ Z�G�. Similarly, by multiplying by h onthe left-hand side, we obtain that h2 ∈ Z�G�. �

Lemma 2.2. If �RG�� is commutative and g, h ∈ G do not commute, then gh =h��g� = ��h�g.

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Proof. As gh �= hg and char�R� �= 2, Lemma 1.2 yields that gh = h��g� orgh = ��h�g.

If gh = h��g�, then Lemma 2.1 gives the result, so assume gh = ��h�g. Hence��h���g� = ��gh� = ��g�h. It follows that ��h� and ��g� satisfy the conditions ofLemma 2.1, so ��h���g� = ����g����h�. Applying � to both sides gives gh = h��g�,as desired. This finishes the proof of the lemma. �

In the above lemma we have shown that for all noncommuting g, h in thegroup G, we have hgh−1 = ��g�. Noncommutative groups with such an involutionhave been described in Goodaire et al. (1996, Theorem III.3.3). They are preciselythe groups with a unique nontrivial commutator and that satisfy the lack ofcommutativity property (“LC” for short). The latter means that for any pair ofelements g, h ∈ G, it is the case that gh = hg if and only if g ∈ Z�G� or h ∈ Z�G�or gh ∈ Z�G�. It turns out (Goodaire et al., 1996, Proposition III.3.6) that suchgroups are precisely those noncommutative groups with G/Z�G� � C2 × C2, whereC2 denotes the cyclic group of order 2. However, for completeness’ sake, we nowinclude an easy proof for the fact that there is a unique nontrivial commutator.

Lemma 2.3. If �RG�� is commutative and both g ∈ G and h ∈ G are not central, theng−1��g� = h−1��h�.

Proof. If gh �= hg, then it follows from Lemma 2.2 that g−1��g� = ��h�h−1.Lemma 1.1 yields that ��h�h−1 = h−1��h� and thus g−1��g� = h−1��h�. Assume nowthat gh = hg and let x ∈ G be so that gx �= xg. If also hx �= xh, then again byLemma 2.2, g−1��g� = x−1��x� = h−1��h�. If on the other hand, hx = xh, theng�xh� �= �xh�g. Again by Lemma 2.2, we obtain that g�xh� = �xh���g� = ���h���x��gand also gx = ��x�g. Hence ��x�hg = ��h���x�g = ��x���h�g and thus h = ��h�.Lemma 1.1 then yields that h ∈ Z�G�, a contradiction. �

If � = ∑g∈G �gg ∈ RG, with �g ∈ R, then we put supp��� = �g ∈ G � �g �= 0�,

this is called the support of �.

Theorem 2.4. Let � be an involution on a nonabelian group G and let R be acommutative ring with char�R� �= 2. The following are equivalent:

1. �RG�� is commutative;2. The group G has the LC property, a unique nontrivial commutator s and the

involution � � G → G is given by

��g� ={g if g ∈ Z�G�

sg if g �∈ Z�G��

3. G/Z�G� � C2 × C2, ��g� = g if g ∈ Z�G� and otherwise ��g� = h−1gh for allh ∈ G with �g h� �= 1.

In this case, �RG�� = Z�RG�. For the classical involution ∗ on G, it follows that �RG�∗is commutative if and only if G is a Hamiltonian 2-group.

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SYMMETRIC ELEMENTS IN GROUP RINGS 731

Proof. Assume that �RG�� is commutative. Let g h ∈ G be such that gh �= hg.Then by Lemma 2.2, we have that 1 �= g−1h−1gh = g−1��g�. Because of Lemma 2.3,it follows that s = g−1��g� is the only nontrivial commutator of G. Since s−1 is alsoa commutator G′ = �1 s� and thus s is central, s2 = 1, ��s� = s, and ��g� = sg forall g ∈ �G\Z�G��.

If g ∈ Z�G� and h �∈ Z�G�, then gh �∈ Z�G�. Thus

�sh���g� = ��h���g� = ��gh� = sgh = shg�

Hence ��g� = g and thus � is as described in the the second part of the statementof the theorem.

To see that G has the LC property, let g h ∈ G be such that gh = hg butg �∈ Z�G� and h �∈ Z�G�. Then ��g� = sg and ��h� = sh. So

��gh� = ��h���g� = shsg = s2hg = gh�

Hence gh ∈ G� and thus, by Lemma 1.1, gh ∈ Z�G�, as desired. This proves (1)implies (2).

We now prove that (2) implies (1). Because G′ = �1 s� we notice that thecenter of RG is generated as an R-module by the center of G and the elements ofthe form g + sg, with g ∈ �G\Z�G�. Let � ∈ RG. Write � = �0 + �1, with �0 �1 ∈ RGand so that supp��0� ⊆ Z�G� and supp��1� ∩ Z�G� = ∅. If � ∈ �RG��, then �0 + �1 =���� = �0 + s�1. So �1 = s�1. It follows that �1 =

∑g �∈Z�G���1�gg and ��1�g = ��1�sg.

Hence �1 and thus also � is central. So we have shown that �RG�� ⊆ Z�RG�, andthus �RG�� = Z�RG�.

As mentioned earlier for the equivalence of (2) and (3) we refer the reader toGoodaire et al. (1996, Proposition III.3.6).

To prove the last part, assume that � is the classical involution ∗ and �RG�∗is commutative. Then from part (2) we get that all elements have order 1, 2,or 4. Further, if g is not a central element, then g2 = s. It follows that everycyclic subgroup of G is normal and thus G is a Hamiltonian 2-group. The resultfollows. �

3. CHARACTERISTIC 2

Throughout this section, R is a commutative ring with identity and char�R� = 2.Further, � is an involution on a group G.

Lemma 3.1. Let g, h ∈ G, and assume �RG�� is commutative. If gh �= hg, then oneof the following holds:

1. gh = ��g���h� and g2 h2 ∈ G�,2. h2 ∈ G� and gh = h��g�,3. g2 ∈ G� and gh = ��h�g.

Proof. Recall from Lemma 1.1 that x��x� = ��x�x ∈ Z�G� for all x ∈ G.Suppose gh �= hg. Then by Lemma 1.2, gh = h��g� or gh = ��h�g or gh =

��g���h� and g2 h2 ∈ G�. If the latter holds then we are in case one of the lemma.

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732 JESPERS AND RUIZ MARÍN

So assume first that gh = h��g�. We show that then h2 = ��h2�. Supposethe contrary. Then, Lemma 1.2 applied to the noncommuting elements gh and hyields that gh2 = h��h���g� or gh2 = ��h�gh. We deal with these cases separately.If gh2 = h��h���g�, then gh = ��g���h�. Hence, gh = h��g� = ��g���h� and thusalso hg = g��h�. On the other hand gh = h��g� implies ��gh� = g��h� and thusgh2 = h��gh� = hg��h� = g��h2�. Consequently, ��h2� = h2, a contradiction. Ifgh2 = ��h�gh, then gh = ��h�g = h��g�. Hence gh2 = ��h�gh = ��h�h��g�. Since��h�h ∈ Z�G�, we thus get that gh2 = ��g���h�h. So gh = ��g���h�. Thereforeh2��g� = h��h�g = ��h�hg = ��h���h���g� = ��h2���g�, and hence h2 = ��h2�,again a contradiction.

Second, assume that gh = ��h�g. We show that then g2 = ��g2�. Ofcourse, the assumption implies that ��h���g� = ��g�����h��. Since ��g� and ��h�do not commute, the previous case applies. Hence ��g�2 = ����g�2� and thusg2 = ��g�2. �

Lemma 3.2. Assume G is nonabelian. If �RG�� is commutative, then one of thefollowing conditions holds:

1. ��g2� = g2 for all g ∈ G,2. A = �g ∈ G � g2 �∈ G� is an Abelian subgroup of index 2 in G such that G� ⊆ A and

��g� = h−1gh, for all h ∈ G\A and g ∈ A.

Proof. Assume that there exists an element x0 ∈ G such that x20 �∈ G�. FromLemma 3.1 it follows at once that A is Abelian. Note that, since by assumption Gis nonabelian, A is a proper subgroup of G.

To show that G� ⊆ A, let x ∈ G� and thus, by Lemma 1.1, x ∈ Z�G�.We claim that �xx0�

2 �∈ G�. Indeed, for otherwise, since x ��x� ∈ Z�G� we getthat ��x0�

2x2 = ��x0�2��x�2 = ���xx0�

2� = �xx0�2 = x2x20 and thus ��x0�

2 = x20, acontradiction. The claim implies that xx0 ∈ A and thus x ∈ A, as desired.

To show that A is a normal subgroup, take a ∈ A with a2 �= ��a2�, g ∈ G\Aand assume that gag−1 �∈ A. Clearly, g2 and �gag−1�2 are in G�, so by Lemma 1.1,ga2g−1 and g2 are central. Consequently, also a2 is central. Thus a2 = ga2g−1 =��ga2g−1� = ��a2�, a contradiction. So A is a normal Abelian subgroup.

Next we show that ��g� = h−1gh for all h ∈ G\A and g ∈ A. For this it isenough to show that ��g� = h−1gh for g ∈ G with g2 �= ��g2�. Let h and g be suchelements. Then of course h2 ∈ G� and Lemma 3.1 yields that gh = hg or gh = h��g�.If gh = hg then, as h �∈ A and g ∈ A, we get that gh �∈ A and thus g2h2 = �gh�2 =���gh�2� = ��g2h2� = ��g2���h2� = ��g2�h2. Therefore ��g2� = g2, a contradiction.So gh = h��g�, that is, ��g� = h−1gh, as desired.

To prove that A has index 2 in G, let g h ∈ G\A. We need to provethat gh−1 ∈ A. Assume the contrary, gh−1 �∈ A. Since x0 ∈ A, the above yields��x0� = �gh−1�−1x0�gh

−1� = h�g−1x0g�h−1 = h��x0�h

−1 = �2�x0�. So ��x0� = x0, acontradiction. �

Theorem 3.3. Let R be a commutative ring with char�R� = 2 and let G be anonabelian group with involution �. If there exists g ∈ G with g2 �∈ G�, then �RG��is commutative if and only if there exist an Abelian subgroup A of index 2 in G andb ∈ G\G� with b2 ∈ G� such that ��a� = b−1ab for all a ∈ A.

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SYMMETRIC ELEMENTS IN GROUP RINGS 733

If �RG�� is commutative, then G� = Z�G� ⊆ A and there exists an element a0 ∈Z�G� of order two so that

��g� ={a0g if g �∈ A

b−1gb if g ∈ A�

Proof. The necessity of the conditions follows from Lemma 3.2. We now provethat they also are sufficient. So, let b ∈ G\G� with b2 ∈ G� and ��a� = b−1ab forall a in an Abelian subgroup A of index 2. Since G is not Abelian, it is clear thatb �∈ A. Let x ∈ G\A. As A has index 2 in G, write x = ab for some a ∈ A. Then, x2 =abab = ab2��a� = a��a�b2. Since b2 a��a� ∈ �G� ∩ A� it follows that also x2 ∈ G�.Hence we have shown that for any x ∈ �G\A� and c ∈ A we have x2 ∈ G� and��c� = x−1cx. In particular, A ∩G� ⊆ Z�G�.

Next we show that G� ⊆ A. Suppose the contrary and let x ∈ �G�\A�. Writex = ab for some a ∈ A. From the above we get x2 = ab2��a� = a��b2���a� andx2 = ��x2� = ��b���a���b���a�. Hence, a��b�=��b���a�=��ab� = ��x� = x = aband thus b = ��b�, a contradiction.

It follows that G� ⊆ Z�G� ∩ A = Z�G�. So, for g ∈ Z�G� we get ��g� =b−1gb = g and thus g ∈ G�. It follows that G� = Z�G�.

To prove that �RG�� is commutative, we thus only have to show that�g + ��g� h+ ��h�� = 0 for all g, h ∈ G. Since A is Abelian, this is obvious ifg, h ∈ A. If g ∈ A and h ∈ G\A, then ��g� = h−1gh and thus h−1�g + ��g��h =��g�+ ����g�� = g + ��g�. Hence h�g + ��g�� = �g + ��g��h and thus also ��h��g + ��g�� = �g + ��g����h�. It follows that indeed �g + ��g� h+ ��h�� = 0.

Let g h ∈ G\A and write h = ag for some a ∈ A. Then, since ��a� = g−1ag,��g2� = g2, g��g� = ��g�g and char�R� = 2, we obtain that

�g + ��g���h+ ��h�� = �g + ��g���ag + ��g���a��

= �g + ��g���g��a�+ ��g���a��

= �g + ��g��2��a�

= 2�g2 + g��g����a�

= 0�

Applying the map � this also yields �h+ ��h���g + ��g�� = 0 and thus�g + ��g� h+ ��h�� = 0. This proves that the mentioned conditions in the statementof the theorem are sufficient.

Finally, we prove that the involution is of the required form. Clearly,��b� = a0b for some 1 �= a0 ∈ A. Hence b = ��b���a0� and thus ��a0� = ��b�−1b =b−1a−1

0 b. But since a0 ∈ A, we also know that ��a0� = b−1a0b. Consequently,a0 = a−1

0 . Furthermore, b2 = ��b�2 = ��b���b� = a0ba0b and thus b = a0ba0. Hence��a0� = b−1a0b = a0. So a0 ∈ G� = Z�G�. Finally, let g = ab be an arbitraryelement in G\A (a ∈ A). Then ��g� = ��b���a� = �a0b��b

−1ab� = a0g. This finishesthe proof. �

Lemma 3.4. Let R be a commutative ring with char�R� = 2. Assume that �RG��is commutative and that g2 ∈ G� for all g ∈ G. Then G′ ⊆ G� ⊆ Z�G�, and G′ andG/Z�G� are elementary Abelian 2-groups. In this case, the involution � � G → G

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734 JESPERS AND RUIZ MARÍN

is given by ��g� = cgg, where cg ∈ G� and c2g = 1. Moreover, if h ∈ G, thencgh = cgch�g h� and if �g h� �= 1, we have that cgh = cg ch or �g h�.

Proof. By Lemma 1.1 G� ⊆ Z�G�. Since, by assumption, g2 ∈ G� ⊆ Z�G� for allg ∈ G, we have that G/G� is an elementary Abelian 2-group. Then G′ ⊆ G�.Moreover, for all g h ∈ G, ghg−1h−1 · g = g2hg−1h−1 = hg2g−1h−1 = hgh−1 and thusghg−1h−1 = hgh−1g−1. Hence �ghg−1h−1�2 = 1 and it follows that G′ is an elementaryAbelian 2-group.

For g ∈ G, write ��g� = cgg. Then, since g2 ∈ G�, we have that cg = ��g�g−1 =

��g−1�g ∈ G�. Since G� ⊆ Z�G� and g2 ∈ G�, we have that c2g = 1. If h ∈ G,then ��gh� = cghgh, while also ��gh� = ��h���g� = chhcgg = cgchhg, and thereforecgh = cgchhgh

−1g−1 = cgch�g h� by an earlier observation. If �g h� �= 1, then byLemma 3.1, one of the following properties holds:

(1) ��g� = h−1gh = hgh−1,(2) ��h� = g−1hg = ghg−1,(3) gh = ��g���h� and g��h� = ��g�h.

If g and h satisfy (1), then cg = �g h� and hence cgh = ch. If g and h satisfy (2), thench = �g h� and hence cgh = cg. And finally if g and h satisfy (3), then cg = ch andhence cgh = �g h�, which finishes the proof of this lemma. �

Theorem 3.5. Let R be a commutative ring with char�R�= 2. Assume that for allg ∈G, g2 ∈ G�. Then �RG�� is commutative if and only if G contains a central subgroupZ such that G/Z is an elementary Abelian 2-group and the involution � � G → G isgiven by ��g� = cgg, where cg ∈ Z and the following properties are satisfied: (1) c2g = 1,(2) ccg = 1, (3) cgh = cgch�g h� and if �g h� �= 1, we have that cgh = cg ch or �g h�.

Proof. The necessity of the conditions follows from Lemma 3.4. Let us prove thatare sufficient. Note that again we get that G′ is an elementary Abelian 2-group. Wehave to prove that for all g h ∈ G, 0 = �g + ��g� h+ ��h�� = gh+ chgh+ cggh+cgchgh− hg − cghg − chhg − cgchhg. If �g h� = 1, then the result follows trivially.So assume that �g h� �= 1. Then, in case cgh = cg, we have that ch = �g h� and theresult follows. In case cgh = ch, we have that cg = �g h� and the result follows. Andfinally, if cgh = �g h�, then cg = ch and again the result holds. �

4. SYMMETRIC UNITS

Throughout this section, we assume that G is a periodic group with aninvolution �, and R is a commutative G-favorable integral domain. Recall (Bovdi,2001) that the ring R is said to be G-favorable if for any g ∈ G of finite order �g�there is a nonzero � ∈ R such that 1− ��g� is invertible in R. Notice that every infinitefield is obviously G-favorable.

The unit group of a ring R is denoted by ��R� and the symmetric units of RGare denoted by ���RG�. So ���RG� = �RG�� ∩��RG�.

Theorem 4.1. Let G be a periodic group. Let R be a G-favorable integral domain.Then the following conditions are equivalent:

1. ���RG� is a subgroup of ��RG�,2. �RG�� is a ring.

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SYMMETRIC ELEMENTS IN GROUP RINGS 735

Proof. Clearly, (2) implies (1). To prove the converse we assume that (1) holds. Inthis case, the symmetric units commute, and thus, of course, G� is a commutativegroup. Let g ∈ G. Since G is a periodic group and R is G-favorable, there is� ∈ R\�0� such that ��g� − 1 is a unit. Then

�g − ��(��g�−1 + ��g�−2g + · · · + �g�g�−2 + g�g�−1

)(1− ��g�

)−1 = 1�

Since �g� = ���g��, it follows that �g − �����g�− �� is a symmetric unit. Since thesymmetric units commute, for all g h ∈ G we have that �g��g� h��h�� = 1 and��g − �����g�− �� h��h�� = 0, and therefore �g + ��g�� and h��h� commute.

Let g h ∈ G\G�, f ∈ G� and let � be nonzero elements in R so that 1− ��g�

and 1− �h� are units of R. Then f �g − �����g�− �� and �h− ����h�− � aresymmetric units.

Since the symmetric units commute and ��g + ��g�� h��h�� = 0 and ��h+��h�� g��g�� = 0, we obtain that ��g + ��g��, �h+ ��h�� and f commute. BecauseR is a domain, � �= 0 and thus the elements g + ��g�, h+ ��h� and f commute.Consequently, the elements in

S = �g + ��g� � g ∈ G\G�� ∪G�

commute. Hence �RG�� is commutative. This proves (2). �

For completeness’ sake, we finish with a comment for the classical involution∗ on integral group rings. In Bovdi and Parmenter (1997) it is proven that if �∗��G�is a group, then the set T�G� of periodic elements in G is a subgroup so that all itssubgroups are normal in G, and T�G� is either Abelian or a Hamiltonian 2-group.These conditions are also sufficient, provided that G/T�G� is a right ordered group.The benefit of working with integral group rings is that quite a number of strongresults on the unit groups ���G� are known. In particular, if u is a nontrivial bicylicunit then Marciniak and Sehgal (1997) have shown that the group �u u∗ is free.Clearly, uu∗, u∗u ∈ �∗��G� but they do not commute. This latter fact is an essentialstep in the proof.

ACKNOWLEDGMENTS

The authors would like to thank the referee for some comments to improvethe article.

The author Eric Jespers has been partially supported by Onderzoeksraad ofVrije Universiteit Brussel, Fonds voor Wetenschappelijk Onderzoek (Vlaanderen),Flemish-Polish bilateral agreement BIL 01/31.

The author Manuel Ruiz Marin has been partially supported by PFMPDI-UPCT-2003 and Fundación Séneca.

REFERENCES

Bovdi, V. (2001). On symmetric units in group algebras. Comm. Algebra 29(12):5411–5422.Bovdi, V., Parmenter, M. M. (1997). Symmetric units in integral group rings. Publ. Math.

Debrecen 50(3–4):369–372.

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Bovdi, V., Kovács, L. G., Sehgal, S. K. (1996). Symmetric units in modular group algebras.Comm. Algebra 24(3):803–808.

Broche Cristo, O. (2003). A Commutatividade dos Elementos Simétricos e Anti-simétricosen Anéis de Grupo, Ph.D. thesis, Univ. Sao Paulo.

Broche Cristo, O. Commutativity of symmetric units in group rings. To appear.Giambruno, A., Sehgal, S. K. (1993). Lie nilpotence of group rings. Comm. Algebra

21:4253–4261.Goodaire, E. G., Jespers, E., Polcino Milies, C. (1996). Alternative Loop Rings. Elsevier.Lee, G. T. (1999). Group rings whose symmetric elements are Lie nilpotent. Proc. Amer.

Math. Soc. 127(11):3153–3159.Marciniak, Z. S., Sehgal, S. K. (1997). Constructing free subgroups of integral group ring

units. Proc. Amer. Math. Soc. 125(4):1005–1009.

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