UNIT GROUPS OF CUBE RADICAL ZERO COMMUTATIVE COMPLETELY PRIMARY FINITE RINGS

UNIT GROUPS OF CUBE RADICAL ZERO COMMUTATIVE COMPLETELY PRIMARY FINITE RINGS

CHITENG’A JOHN CHIKUNJI Received 1 July 2004

A completely primary finite ring is a ringRwith identity 1=0 whose subset of all its zero divisors forms the unique maximal idealJ. LetRbe a commutative completely primary finite ring with the unique maximal idealJ such thatJ³=(0) andJ²=(0). ThenR/J^∼= GF(p^r) and the characteristic ofRisp^k, where 1≤k≤3, for some primepand positive integerr. LetRo=GR(p^kr,p^k) be a Galois subring ofRand let the annihilator ofJbeJ²so thatR=Ro⊕U⊕V, whereUandVare finitely generatedRo-modules. Let nonnegative integerssandtbe numbers of elements in the generating sets forUandV, respectively.

Whens=2,t=1, and the characteristic ofRisp; and whent=s(s+ 1)/2, for any fixed s, the structure of the group of unitsR^∗of the ringRand its generators are determined;

these depend on the structural matrices (a_{i j}) and on the parametersp,k,r, ands.

Notations

Throughout this paper,Rwill denote a finite ring, unless otherwise stated,Jwill denote the Jacobson radical ofR, and we will denote the Galois ringGR(p^nr,pⁿ) of characteristic pⁿand orderp^nrbyR_o, for some primep, and positive integersn,r.

We denote the group of units ofRbyR^∗and a cyclic group of orderπby(π). Ifg is an element ofR^∗, theno(g) denotes its order, andgdenotes the cyclic group generated byg. Furthermore, for a subsetAofRorR^∗,|A|will denote the number of elements inA.

The ring of integers modulo the numbernwill be denoted byZn, and the characteristic ofRwill be denoted by charR.

1. Introduction

In [6], Fuchs asked for a characterization of abelian groups which could be groups of units of a ring. This question was noted to be too general for a complete answer [12], and a natural course is to restrict the classes of groups or rings to be considered.

LetRbe a ring and letR^∗denote its multiplicative group of unit elements. All local ringsRwithR^∗cyclic were determined by Gilmer [8] and this case was also considered by Ayoub [1] (also proofs are given in [10,11]). Pearson and Schneider have found all

Copyright©2005 Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences 2005:4 (2005) 579–592 DOI:10.1155/IJMMS.2005.579

RwhereR^∗is generated by two elements. Clark [4] has investigatedR^∗where the ideals form a chain and has shown that ifp≥3,n≥2, andr≥2, then the units of the Galois ringGR(p^nr,pⁿ) are a direct sum of a cyclic group of orderp^r−1 andrcyclic groups of orderpⁿ−1 (this was also done independently by Raghavendran [11]). In fact, Raghaven- dran described the structure of the multiplicative group of every Galois ring. Stewart in [12] considered a related problem to that asked by Fuchs [6] by proving that for a given finite groupG(not necessarily abelian), there are, up to isomorphism, only finitely many directly indecomposable finite rings having group of units isomorphic toG.

Ganske and McDonald [7] provided a solution forR^∗when the local ringRhas Jacob- son radicalJsuch thatJ²=(0) by showing that

R^∗=

⊕ nt i=1

(p)

⊕

|K| −1, (1.1)

wheren=dimK(J/J²),|K| =p^t, and(π) denotes the cyclic group of orderπ.

In [5], Dolzan found all nonisomorphic rings with a group of units isomorphic to a groupGwithnelements, wherenis a power of a prime or any product of prime powers, not divisible by 4; and also found all groups withnelements which can be groups of units of a finite ring, a contribution to Stewart’s problem [12]. More recently, X.-D. Hou et al.

gave an algorithmic method for computing the structure of the group of units of a finite commutative chain ring and further strengthening the known result by listing a set of linearly independent generators for the group of units.

The present paper focuses on the group of unitsR^∗of a commutative completely pri- mary finite ringR with unique maximal idealJ such thatR/J^∼₌GF(p^r),J³=(0), and J²=(0) so that the characteristic ofRisp^k, where 1≤k≤3; and further identifies sets of generators forR^∗.

In particular, letRo=GR(p^kr,p^k) be a Galois subring ofRand let the annihilator of J beJ² so thatR=Ro⊕U⊕V, whereU andV are finitely generatedRo-modules. Let nonnegative integerssandtbe numbers of elements in the generating sets forUandV, respectively. Whens=2,t=1, and charR=p, and whent=s(s+ 1)/2, for any fixeds, the structure of the group of unitsR^∗of the ringRand its generators have been determined;

these depend on the structural matrices (a_{i j}) and on the parametersp,k,r, ands.

2. Preliminaries

We refer the reader to [2] for the general background of completely primary finite rings Rwith maximal idealsJ such thatJ³= {0}andJ²= {0}. LetRbe a completely primary finite ring with maximal idealJ such thatJ³=(0) andJ²=(0). ThenRis of order p^nr and the residue fieldR/Jis a finite fieldGF(p^r), for some primepand positive integersn, r. The characteristic ofRisp^k, wherekis an integer such that 1≤k≤3. LetGR(p^kr,p^k) be the Galois ring of characteristic p^k and order p^kr, that is,GR(p^kr,p^k)=Zp^k[x]/(f), where f ∈Zp^k[x] is a monic polynomial of degreerwhose image inZp[x] is irreducible.

Then, it can be deduced from the main theorem in [4] thatRhas a coefficient subringR_o of the formGR(p^kr,p^k) which is clearly a maximal Galois subring ofR. Moreover, there

exist elementsm1,m2,. . .,m_h∈Jand automorphismsσ1,. . .,σ_h∈Aut(R_o) such that R=R_o⊕

h i=1

R_om_i (2.1)

(asRo-modules),mir=r^σimi, for everyr∈Roand anyi=1,. . .,h. Further,σ1,. . .,σhare uniquely determined byRandRo. The maximal ideal ofRis

J=pRo⊕ h i=1

Romi. (2.2)

It is worth noting thatRcontains an elementb of multiplicative orderp^r−1 and that R_o=Zp^k[b] (see, e.g., [2, Result 1.3]).

The following results will be useful.

Proposition2.1. LetRbe a completely primary finite ring (not necessarily commutative).

Then the group of unitsR^∗ofRcontains a cyclic subgroupbof orderp^r₋1, andR^∗is a semidirect product of1 +Jandb.

Proof. Obviously, the group of unitsR^∗ of R isR−J, |R^∗| =p^(n−1)r(p^r−1), and φ: R→R/Jinduces a surjective multiplicative group homomorphismϕ:R^∗→(R/J)^∗. Since kerφ=J, we have kerϕ=1 +J. In particular, 1 +Jis a normal subgroup ofR^∗.

Letβ =(R/J)^∗, and letb_o∈ϕ⁻¹(β). Then, the multiplicative order ofb_ois a multiple of p^r−1 and a divisor of|R−J| =p^nr−p^(n−1)r=p^(n−1)r(p^r−1); hence, of the form p^s(p^r−1). But thenb=bo^pshas multiplicative orderp^r−1 andϕ(bo^ps)=β^ps, which is still a generator of (R/J)^∗, since (p^s,p^r−1)=1.

Finally, since|R^∗| = |1 +J| · |b|, and (1 +J)∩ b =1, we haveR^∗=(1 +J)· b,

hence,R^∗=(1 +J)×θb, a semidirect product.

Proposition2.2. LetRbe a completely primary finite ring (not necessarily commutative).

Then the group of unitsR^∗is solvable.

Proof. ThatR^∗is a solvable group follows from the fact that 1 +Jis a normalp-subgroup

ofR^∗, andR^∗/(1 +J) is cyclic.

Lemma2.3. LetRbe a completely primary finite ring (not necessarily commutative). IfGis a subgroup ofR^∗of orderp^r−1, thenGis conjugate tobinR^∗.

Proof. This follows from key properties ofp-solvable groups contained in the variation of Sylow’s theorem, due to Philip Hall, since the order ofGis prime to its index inR^∗(see,

e.g., [9, Theorem 8.2 page 25]).

Proposition2.4. LetRbe a completely primary finite ring (not necessarily commutative).

IfR^∗contains a normal subgroup of orderp^r−1, then the setKo= b ∪ {0}is contained in the center of the ringR.

Proof. ByLemma 2.3,bis normal inR^∗and since 1 +J is a normal subgroup ofR^∗ with|b ∩(1 +J)| =1, it follows thatband 1 +Jcommute elementwise. Hence,blies

in the center ofR.

Proposition 2.5. Let Rbe a completely primary finite ring. Then, (1 +Jⁱ)/(1 +Jⁱ⁺¹)^∼= Jⁱ/Jⁱ⁺¹(the left-hand side as a multiplicative group and the right-hand side as an additive group).

Proof. Consider the map

η:1 +Jⁱ/1 +Jⁱ⁺¹−→Jⁱ/Jⁱ⁺¹ (2.3) defined by

(1 +x)1 +Jⁱ⁺¹−→x+Jⁱ⁺¹. (2.4)

Then it is easy to see thatηis an isomorphism.

Remark 2.6(see [3, Result 2.7]). LetRbe a completely primary finite ring of characteristic p^k and with Jacobson radicalJ. LetR_o be a Galois subring ofR. Ifm∈J and p^t is the additive order ofm, for some positive integert, then|Rom| =p^tr.

Proof. Apply the fact that

Rom^∼=Ro/ p^tRo. (2.5)

Now letRbe a commutative completely primary finite ring with maximal idealJsuch thatJ³=(0) andJ²=(0). In [2], the author gave constructions describing these rings for each characteristic and for details, we refer the reader to [2, Sections 4 and 6].

IfRis a commutative completely primary finite ring with maximal idealJ such that J³=(0) andJ²=(0), then from Constructions A and B [2],

R=Ro⊕U⊕V⊕W, (2.6)

J=pR_o⊕U⊕V⊕W, (2.7)

where theRo-modulesU,V, andWare finitely generated. The structure ofRis charac- terized by the invariantsp,n,r,d,s,t, andλ; and the linearly independent matrices (a^k_{i j}) defined in the multiplication. Let ann(J) denote the two-sided annihilator ofJinR. No- tice that sinceJ²⊆ann(J), we can writeR=R_o⊕U⊕M, and hence,J=pR_o⊕U⊕M, whereM=V⊕W, and the multiplication inRmay be written accordingly. It is therefore easy to see that the description of rings of this type reduces to the case where ann(J) co- incides withJ². Therefore, when investigating the structure of the group of units of this type of rings for a given order, say p^nr, where ann(J) does not coincide withJ², we will first write all the rings of this type of order≤p^nr, where ann(J) coincides withJ².

In what follows, we assume that ann(J)=J².

LetRo=GR(p^kr,p^k)(1≤k≤3) and let nonnegative integerssandtbe numbers of ele- ments in the generating sets{u1,. . .,us}and{v1,. . .,vt}for finitely generatedRo-modules U and V, respectively, wheret≤s(s+ 1)/2. Assume thatu1,u2,. . .,u_s andv1,. . .,v_t are commuting indeterminates. ThenR=Ro⊕U⊕V.

ByProposition 2.1, and sinceRis commutative,

R^∗= b ·(1 +J)^∼₌ b ×(1 +J), (2.8) a direct product.

Again, notice that sinceRis of orderp^nrandR^∗=R−J, it is easy to see that|R^∗| = p^(n−1)r(p^r−1) and |1 +J| = p^(n−1)r, so that 1 +J is an abelian p-group. Thus, R^∗∼₌ (abelianp-group)×(cyclic group of order|R/J| −1).

Our goal is to determine the structure and identify a set of generators of the multi- plicative abelianp-group 1 +J.

3. The group1 +J

Now letRbe a commutative completely primary finite ring with maximal idealJ such thatJ³=(0) andJ²=(0). Let 1 +Jbe the abelianp-subgroup of the unit groupR^∗.

The group 1 +Jhas a filtration 1 +J⊃1 +J²⊃1 +J³= {1}with filtration quotients (1 +J)/(1 +J²) and (1 +J²)/{1} =1 +J² isomorphic to the additive groupsJ/J² andJ², respectively.

Remark 3.1. Notice that 1 +J²is a normal subgroup of 1 +J. But, in general, 1 +Jdoes not have a subgroup which is isomorphic to the quotient (1 +J)/(1 +J²) as may be illustrated by the following example.

Example 3.2. LetR=Zp³, where p is an odd prime. ThenJ=pZp³, ann(J)=J², and 1 +J^∼₌Zp², 1 +J^2∼₌Zp, (1 +J)/(1 +J²)^∼=Zp.

Remark 3.3. In view of the above remark and example, we investigate the structure of 1 +Jby considering various subgroups of 1 +J.

3.1. The case whens=2,t=1, andcharR=p. Supposes=2,t=1, and charR=p. Let Ro=Fq=GF(p^r), the Galois field ofq=p^relements. Then

R=Fq⊕Fqu1⊕Fqu2⊕Fqv, (3.1) the Jacobson radical

J=Fqu1⊕Fqu2⊕Fqv, (3.2)

J²=Fqv. (3.3)

The multiplication inRis given by

u²₁₌a11v, u1u2=u2u1=a12v, u²₂₌a22v, (3.4) whereai j∈Fq. The elementsai jform a nonzero symmetric matrix

a11 a12

a21 a22

(3.5) sinceJ²=(0).

SinceR^∗is a direct product of the cyclic groupbof orderp^r−1 and the group 1 +J of orderp^3r, it suffices to determine the structure of 1 +J.

In this case,

1 +J=1 +Fqu1⊕Fqu2⊕Fqv, (3.6) and sincesandtare fixed, the structure of 1 +Jnow depends on the primep, the integer r, and the structural matrix^aa¹¹21 ^aa¹²22

. We investigate this by considering cases depending on the type of the structural matrix.

Letε1,ε2,. . .,ε_r be elements ofFq withε1=1 so thatε1,ε2,. . .,ε_r form a basis forFq

regarded as a vector space over its prime subfieldFp. Case (i). Suppose that^aa¹¹21^aa¹²22

=_a0

0 0

, witha=0. Then

1 +J^∼=



Z^r4×Z^r2, if charR=2,

Z^r_p×Z^r_p×Z^r_p, if charR=p=2. (3.7) To see this, we consider the two cases separetely. So, suppose thatp=2. We first note the following results:

1 +εiu1∈1 +J, 1 +εiu1

4

=1, 1 +εiu2

2

=1, g⁴=1,∀g∈1 +J. (3.8) For positive integersk_i,l_i, withk_i≤4,l_i≤2, we assert that

r i=1

1 +εiu1

ki

·

r i=1

1 +εiu2

li

=1 (3.9)

will implyk_i=4 for alli=1,. . .,r; andl_i=2 for alli=1,. . .,r.

If we setFi= {(1 +εiu1)^k|k=1,. . ., 4}for alli=1,. . .,r; andGi= {(1 +εiu2)^l|l=1, 2} for alli=1,. . .,r, we see thatFi,Gi are all cyclic subgroups of the group 1 +J and that these are of the precise orders indicated by their definition. The argument above will show that the product of 2rsubgroupsF_iandG_iis direct. So, their product will exhaust the group 1 +J.

Whenpis an odd prime, we have to consider the equation

r i=1

1 +εiu1

ki

·

r i=1

1 +εiu2

li

·

r i=1

1 +εiv^mi=1 (3.10)

and as each element in 1 +J raised to the powerpequals 1, we see that 1 +Jwill be an elementary abelian group.

Case (ii). Suppose that^aa¹¹21 ^aa¹²22

=_0a

a0

, witha=0. Then

1 +J^∼₌Z^r_p×Z^r_p×Z^r_p, (3.11)

for everyp=charR. In this case, we consider the equation

r i=1

1 +εiu1

ki

·

r i=1

1 +εiu2

li

·

r i=1

1 +εiv^mi=1 (3.12)

and the integersk_i,l_i,m_iwill implyk_i=l_i=m_i=pfor alli=1,. . .,r.

If we setFi= {(1 +εiu1)^k|k=1,. . .,p}for alli=1,. . .,r;Gi= {(1 +εiu2)^l|l=1,. . .,p} for alli=1,. . .,r; andHi= {(1 +εiv)^m|m=1,. . .,p}for alli=1,. . .,r, we see thatFi,Gi, andH_iare all cyclic subgroups of the group 1 +Jand that these are all of order p. The product of the 3r subgroupsFi,Gi, andHi is direct. So, their product will exhaust the group 1 +J.

Case (iii). Suppose now that^aa¹¹21 ^aa¹²22

=_{a b}

b0

, withaandbbeing nonzero. Then

1 +J^∼₌



Z^r4×Z^r2, if charR=2,

Z^r_p×Z^r_p×Z^r_p, if charR=p=2. (3.13) The argument is similar to that in Case (i).

Case (iv). Suppose^aa¹¹21 ^aa¹²22

=_a0

0b

, withaandbbeing nonzero. Thenu²1=av,u²2=bv, andu1u2=u2u1=0.

If charR=p=2, theno(1 +εiu1)=o(1 +εiu2)=p(i=1,. . .,r). Moreover, for every i=1,. . .,r,1 +εiu1 ∩ 1 +εiu2 = {1}. Also,o(1 +εiv)=p, and the element 1 +εiv(i= 1,. . .,r) generates a cyclic subgroup of orderp.

If charR=2, then in 1 +J, we see thato(1 +εiu1)=4 and for eachεi, by considering the element 1 +εiu1+εiu2+εivof order 2, one obtains the direct product

1 +J=

r i=1

1 +εiu1

×

r i=1

1 +εiu1+εiu2+εiv. (3.14)

Hence,

1 +J^∼=



Z^r4×Z^r2, if charR=2,

Z^r_p×Z^r_p×Z^r_p, if charR=p=2. (3.15) Case (v). Finally, suppose that^aa¹¹21^aa¹²22

=_{a b}

b c

, witha,b, andc being nonzero. Then u²1=av,u²2=cv, andu1u2=u2u1=bv. In this case, it is easy to verify that

1 +J^∼₌



Z^r4×Z^r2, if charR=2,

Z^r_p×Z^r_p×Z^r_p, if charR=p=2. (3.16) The number of cases involved in determining the structure of 1 +Jfor larger values of sand fort < s(s+ 1)/2 compels us to investigate the problem by considering the extreme case when the invariantt=s(s+ 1)/2, and to leave the other cases for subsequent work.

3.2. The case whent=s(s+ 1)/2, forsfixed. Suppose thatt=s(s+ 1)/2 for a fixed non- negative integer s. Let u1,u2,. . .,u_s be commuting indeterminates over the Galois ring Ro=GR(p^kr,p^k), where 1≤k≤3. Then it is easy to verify that

R=Ro⊕ s i=1

Roui⊕ s i,j=1

Rouiuj, (3.17)

where

uiuj=ujui, u³_i =u²_iuj=uiu²_j=0, for everyi,j=1,. . .,s, (3.18) is a commutative completely primary finite ring with Jacobson radical

J=pRo⊕ s i=1

Roui⊕ s i,j=1

Rouiuj; (3.19)

J²=pRo⊕ s i,j=1

Rouiuj or J²=p²Ro⊕ s i,j=1

Rouiuj; J³=(0). (3.20) In this case, the linearly independent matrices (a^k_{i j}) defined in the multiplication ofR are thet=s(s+ 1)/2,s×ssymmetric matrices with 1’s in the (i,j)th and (j,i)th positions, and zeros elsewhere.

It follows clearly that

1 +J₌1 +pR_o⊕ s i=1

R_ou_i⊕ s i,j=1

R_ou_iu_j, (3.21)

and it can easily be deduced that every elementxof 1 +Jhas a unique expression of the form

x=1 +pa_o+ s i=1

a_iu_i+ s i,j=1

a_{i j}u_iu_j, (3.22)

whereao,ai,ai j=ajiare inK=Ro/ pRo.

Letsbe a fixed nonnegative integer and suppose thatt=s(s+ 1)/2. If charR=p, then

|R| =p^{((s2+3s+2)/2)r}, |J| =p^((s2+3s)/2)r (3.23) because|R_ou_i| =p^r(for eachi=1,. . .,s) and|R_ou_iu_j| =p^r(fori,j=1,. . .,s); thus

|1 +J| =p^((s2+3s)/2)r. (3.24)

If charR=p², then

|R| =p^{((s2+5s+4)/2)r}, |J| =p^{((s2+5s+2)/2)r} (3.25)

because|R_o| =p^2r,|pR_o| =p^r,|R_ou_i| =p^2r, ifpu_i=0 (for eachi=1,. . .,s) and|R_ou_iu_j|

=p^r(fori,j=1,. . .,s) (seeRemark 2.6), and thus

|1 +J| =p^{((s2+5s+2)/2)r}. (3.26)

Finally, if charR=p³, then

|R| =p^{((s2+5s+6)/2)r}, |J| =p^{((s2+5s+4)/2)r} (3.27) because|Ro| =p^3r,|pRo| =p^2rand ifpui=0,|Roui| =p^2r (becausep²ui=0) (for each i=1,. . .,s) and|Rouiuj|=p^r(fori,j=1,. . .,s) (seeRemark 2.6and also becausepuiuj=0), and hence,

|1 +J| =p^{((s2+5s+4)/2)r}. (3.28)

Proposition3.4. IfcharR=p^k, wherek=2or3, then1 +Jcontains1 +pRoas its sub- group.

Proof. We only show the case for charR=p², the other case follows easily from this. Now, each element of 1 +pR_ois of the form 1 +pr, for everyr∈R_o, and for any two elements 1 +pr1and 1 +pr2, we have

1 +pr1

1 +pr2

=1 +pr1+r2

(3.29)

which is clearly an element of 1 +pR_o.

Proposition3.5. For each pairu_i,u_jwithi=jandu_iu_j=u_ju_i,1 +R_ou_iu_jis a subgroup of1 +J.

Proof. It is easy to see that 1 +Rouiujis a subgroup of 1 +Jbecause for any two elements 1 +r1uiujand 1 +r2uiujin 1 +Rouiuj, we have

1 +r1u_iu_j1 +r2u_iu_j=1 +r1+r2

u_iu_j∈1 +R_ou_iu_j (3.30)

since (u_iu_j)²=0.

Proposition3.6. For everyi=1,. . .,s,1 +R_ou_i+R_ou²_i is a subgroup of1 +J.

Proof. Obviously,

1 +r1u_i+r2u²_i1 +s2u_i+s2u²_i=1 +r1+s1

u_i+r1s1+r2+s2

u²_i (3.31)

lies in 1 +Roui+Rou²_i, for any pair 1 +r1ui+r2u²_i and 1 +s2ui+s2u²_i of elements in 1 +

R_ou_i+R_ou²_i.

In view of Remark 2.6 and Propositions 3.4, 3.5, and 3.6, we may now state the following.

Proposition3.7. Let1 +pR_o,1 +R_ou_i+R_ou²_i, and1 +R_ou_iu_j be the subgroups of1 +J defined above. Then

1 +pRo=





p^r, if charR=p²,

p^2r, if charR=p³, (3.32) 1 +R_ou_i+R_ou²_i=











p^2r, if charR=p, p^3r, if charR=p², p^3r, if charR=p³,

(3.33)

1 +R_ou_iu_j=p^r, (3.34)

for every characteristic ofR.

Proposition3.8. The group1 +Jis a direct product of the subgroup1 +pRo,ssubgroups 1 +Roui+Rou²_i, ands(s−1)/2subgroups1 +Rouiuj, wherei=janduiuj=ujui.

Proof. This follows from the fact that 1 +pRo, 1 +Roui+Rou²_i, and 1 +Rouiuj are sub- groups of 1 +J, intersection of any pair of these subgroups is trivial (for everyi, j= 1,. . .,s), and byProposition 3.7,

|1 +J| =1 +pR_o×

s i=1

1 +R_ou_i+R_ou²_i×

s i=j=1

1 +R_ou_iu_j. (3.35) 3.2.1. The structure of1 +pRo. The structure of 1 +pRo is completely determined by Raghavendran in [11]. For convenience of the reader, we state here the results useful for our purpose. For detailed proofs, refer to [11, Theorem 9].

We takerelementsε1,. . .,εr inRowithε1=1 such that the set{ε1,. . .,εr}is a basis of the quotient ringR_o/ pR_oregarded as a vector space over its prime subfieldGF(p). Then we have the following.

Proposition3.9 [11, Theorem 9]. IfcharRo=p², then1 +pRois a direct product ofr cyclic groups1 +pεj, each of orderp, for any primep.

Proposition3.10 [11, Theorem 9]. Let charRo=p³. If p=2, then1 +pRo is a direct product of2cyclic groups−1 + 4ε1and1 + 4ε1, each of order2, and(r−1)cyclic groups 1 + 2εj(j=2,. . .,r), each of order4. If p=2, then1 +pRo is a direct product ofrcyclic groups1 +pεj(j=1,. . .,r), each of orderp².

3.2.2. The structure of1 +Roui+Rou²_i. We now consider the structure of the subgroup 1 +Roui+Rou²_i of the p-group 1 +J. We first note that if charRo=p, thenRo=GF(p^r) the field of p^r elements, if charR_o=p², thenR_ois the Galois ringGR(p^2r,p²) of order p^2r, and if charRo=p³,Ro=GR(p^3r,p³) the Galois ring of orderp^3r.

We chooserelementsε1,. . .,εrinRowithε1=1 such that the set{ε1,. . .,εr}is a basis of the quotient ringR_o/ pR_oregarded as a vector space over its prime subfieldGF(p). Then we have the following.

Proposition3.11. LetcharR_o=p. If p=2, then1 +R_ou_i+R_ou²_i is a direct product ofr cyclic groups1 +ε_ju_i(j=1,. . .,r), each of order4. Ifp=2, then1 +R_ou_i+R_ou²_i is a direct product of2rcyclic groups1 +εjuiand1 + 2εjui(j=1,. . .,r), each of orderp.

Proof. If charRo=2, then1 +εjui is of order 4, for every j=1,. . .,rand for anyi= 1,. . .,s, and hence

r j=1

1 +ε_ju_i=4^r=2^2r=1 +R_ou_i+R_ou²_i. (3.36)

Therefore, the product^r_j=11 +εjuiis direct.

Similarly, if charRo=p=2, the elements 1 +εjuiand 1 + 2εjuiare each of orderp, 1 +εjui

∩

1 + 2εjui

= {1}, (3.37)

for everyj=1,. . .,r, and

r j=1

1 +ε_ju_i·

r j=1

1 + 2ε_ju_i=p^r·p^r=p^2r=1 +R_ou_i+R_ou²_i, (3.38)

hence

1 +R_ou_i+R_ou²_i =

r j=1

1 +ε_ju_i×

r j=1

1 + 2ε_ju_i, (3.39)

a direct product.

Proposition3.12. LetcharRo=p². Ifp=2, then1 +Roui+Rou²_i is a direct product ofr cyclic groups1 + 2εjui, each of order2, andrcyclic groups1 + 3εjui(j=1,. . .,r), each of order4. Ifp=2, then1 +R_ou_i+R_ou²_i is a direct product ofrcyclic groups1 +pε_ju_i, each of orderp, andrcyclic groups1 +ε_ju_i(j=1,. . .,r), each of orderp².

Proof. Suppose charR_o=p². Ifp=2,1 + 2ε_ju_iis of order 2 and1 + 3ε_ju_iis of order 4, 1 + 2εjui

∩

1 + 3εjui

= {1}, (3.40)

for everyj=1,. . .,rand anyi=1,. . .,s. Since

r j=1

1 + 2εjui·

r j=1

1 + 3εjui=2^r·4^r=2^3r=1 +Roui+Rou²_i, (3.41)

it follows that

1 +Roui+Rou²_i =

r j=1

1 + 2εjui

×

r j=1

1 + 3εjui

(3.42)

is a direct product.