On the balanced subgroups of modular group rings

Let S ( RG) be the normed p-unit group in a group ring RG, formed by an abelian group G and a commutative ring R with identity of prime characteristic p . All unexplained symbols and letters as well as the terminology and definitions from the abelian group theory (including the topological ones) can be found in the classical book monographs [7]. For a background material in that direction, we refer the reader also to [1]–[6].

The major goal motivating the present paper is to find some special nice and isotype subgroups of S ( RG), a problem that arises naturally in the examination of the total projectivity both in modular and semi-simple aspects (cf. [1] and [6]). Thus the property of subgroups being balanced in modular group rings is crucial for the investigation of nice composition series and nice bases in such rings (see, for instance, [9] or [5]).

Moreover, the balanced subgroups play an important role for the quasi-completeness (e. g. [2, 3]) and torsion-completeness (e. g. [4]) in group algebras by using either an algebraical or topological technique in terms of bounded convergent Cauchy sequences.

The query for the balanced property of S ( KH ) in S ( KG) when KG is semisimple, such that G is p -primary and K is either a field having arbitrary characteristic or is a special ring of zero characteristic, is considered and settled in some way by us in [4].

In [9] and [8], May and Hill-Ullery studied the case when R is a field, whereas we here investigate the general situation which cannot be treated by similar reasons.

The main result

We start with a single key assertion needed for future applications. It discovers the balanced property in S ( RG) of subgroups of the type S ( RH ), whenever H 6 G; for certain other balanced subgroups the readers can see [6].

Proposition. Let H be a p -balanced ( that is p -nice and p -isotype ) subgroup of G . Then S ( PH ) is balanced in S ( RG) , provided P is a perfect subring of R with the same unity.

C « p -nice» . Bearing in mind [7], it is enough to calculate that Q [S p ^a ( RG)S ( PH )] = α<τ

S ^p ( RG)S ( PH ) for every limit ordinal т . In fact, given an element x in the left hand-side,

mnm hence, by [3], x G (P rigi)S (PH) = (P ri gi )S (PH) = ..., where ri £ Rp, P ri = 1, i=1                     i=1

α          βn gi G Gp ; ri G Rp , P ri = 1, gi G Gp ; a < в < т and в is arbitrary but a fixed ordinal. i=1

Thus we can write

m rigi = i=1

Xn ri0gi0    Xn fihi i=1         i=1

= E Er i f g i h j , ij

n whenever fi G P with   fi = 1 and hi G H .

i =1

Writing   ri0 fj gi0 hj in canonical form, we may presume without loss of generality that the i,j following relations hold:

0       0         0       0       00         0

^r i J i = ⁰, ^r 1 ^f 2 = • • • = ^r i J n = ^{0; r} 2 ^f 2 = ⁰, ^r 2 J i = ^r 2 ^f 3 = • • • = ^r 2 f n = ^0; • • • ^;

^r S ^f s = °, ^r S ^f i = • • • = ^r S ^f s- i = ^r S ^f s +1 = • •• = ^r S ^f n = 0

for some s ∈ N , and all other ring products are not zero. Of course, these ring dependencies are indeed correct and well-chosen, because if in addition rf i = 0 we detect that 0 = r ^ f + • • • + f n ) = r i which is a contradiction. Moreover, we note that r i f i = r i (f i + • • • + f n ) = r ( , • • •, r S f s = r S (f i + • • • + f n ) = r S .

Now, let us assume for difficulty that the following additional group ratios hold (if not, the things are easy): g ₂ ⁰ h 2 = g ₃ ⁰ h 3 = • • • = g _S ⁰ _{- 1} h _{S -} 1 such that r ₂ ⁰ f 2 + r ₃ ⁰ f 3 + • • • + r _S ⁰ _{- 1} f _{S -} 1 = 0, i. e. these elements do not lie in the support.

A crucial fact is that, since the supports of the elements in the group ring are finite while the set { a < в < т : в >  ^ш } is infinite, all given relations are assumed of the above types presented. We mention that all other variants, even when there is no zero divisors, are identical or have a simple interpretation.

The canonical records imply ri = rf gi = g0hi; r2 = rS+iA, g2 = gS+ihi; r3 = rS+2/2, дз = gS+2h2; • • •;

r k = ^r S +i ^f 2 , g k = g S +i h 2 r k +i = ^r S +2 Л, g k +i = g S +2 h i ; • • •; r S = ^r S ^f s , g S = g S h s •• •;

^r n = ^r _n ^f n , ^g n = ^g _n ^h n ^; • • • ^{; r} m- 2 = ^r _{n -} 2 ^f n- i , ^g m- 2 = ^g _{n -} 2 ^h n- i ^;

^r m- i = ^r _{n -} i ^f n i ^g m- i = ^g _{n -} i ^h n ^{; r} m = ^r _n^fl i ^g m = g n h i ^

Therefore, we get that, ri G T Rpe = RpT, • • •, rm G RpT, hence ri G RpT, • • •, гП G RpT since β<τ r'i = r(fi = ri, • ••, rS = rS/s = rS, rS0 +1 = rS0 +1f1 + • • • + rS0 +1fn = r2 + • • • ,

..., rn = rn f1 + • • • + rn fn = rm + • • • + rn , where m = n2 — s + 2 — s(n — 1) = n2 — sn + 2. Besides, gi G \ (Gpe H) = GpT H, •••,gm G GpT H^

β<τ

Thus we can write g i = g _Ti a i , ..., g m = g _Tm a _m where g _T i ,..., g _Tm E G ^{p T} and a i ,..., a m E H . Since g i g ^{- i} E G ^{p T} , whence a i a ^{- i} E G ^{p T} , we shall presume that a i = a ^ because g _Ti a i = g T 022 for some g T _i E G p . By the same token we may produce also for the other pairs of indices ( i, j ) such that g i g^-1 E G p . Besides, g i g ^- = h i h ^- E H , hence g _Ti g - E H . The same procedure can be done for the other pairs of distinct indexes with this property as well.

mnn

We observe that 52 rigi = I 52 rigTui I 52 fibi , where for 1 6 i 6 n we have bi = aui i=i           i=ii=i or so

bi = auigTvigTWi E H for some appropriate permutations Ui, Vi, Wi of the indexes 1,..., n that gT2b2 = gTзЬз = • • •

= д т ( s- i) b s- i , and eventually r i = r u i .

mm

When m > n it may be possible that 52 rigi = I 52 rigTi ) a for some a E H• i=i        Vi=i/

m

Since 52 r i g i E S(RG), there exists a group member from the sum which member belongs i =1

nn to Gp . By a reason of symmetry the same should be valid even for    ri0 gi0 and    fi hi . So, i=i           i=i with no harm of generality, we may suppose that: gi,... ,gi E Gp, ri + ... + ri — 1 belongs to the nil-radical of R; Gp 3 gi+i E gi+2GP E ... E gmGP, ri+i + ri+2 + • • • + rm lies in the nil-radical of R; l E N. Analogously gi,... ,g'k E Gp, ri + ... + rk — 1 belongs to the nil-radical of R; Gp 3 gk+i E gk+ 2Gp E ... E g'nGp, rk+i + rk+2 + ... + rn lies in the nilradical of R and hl,... ,hk E Hp, fi + ... + fk — 1 is in the nilradical of R; Hp 3 hk+i E hk+2 Hp E ... E hnHp, fk+i + fk+2 + ... + fn is in the nilradical of R; n E N.

Because, for any ordinal 5, we know that ( G p H ) _p = G p H p , we will presume that g _{T i} E G ^pp ^T and a 1 E H p . Moreover, by what we have already proved,

ττ       τ       τ gl+igl+2 E (G H)P = Gp HP, . . . , gl+igm E Gp Hp, . . . , gl+2gm E Gp Hp etc. gk+1gk+2 E Gp Hp , . . . , gk+1gn E Gp Hp , . . . , gk+2gn E Gp Hp etc.

Similarly for h _k+i h ^- +₂ E H _p , ..., h _k+i h^-1 E H _p , ..., h _k+2 h^-1 E H _p etc.

Furthermore, b i = a u i g _Tv i g ^W i E H _p for i = 1,... ,k or b i E b j H _p for k + 1 6 i = j 6 n.

Finally, it is apparent that PP r 0 g _Tu i E T S ( R^{p< a} G p " ) = S ( R ^p G p ) = S ^{p T} (RG) and i =i           а<т

PP f i b i E S ( PH ). That is why, it is easily checked that x E S ^{p T} ( RG ) S ( PH ). Thereby, the i=1

wanted equality is true, as expected.

« p -isotype» . Exploiting [1],

S ( PH ) П S ^p “ (RG) = S ( PH ) П S ( R^P“G^p“ ) = S (P(H П G ^p “ )) = S ( PH ^p “ ) = S ^p “ ( PH ) .

So, the proof is completed in all generality. B