Publicacions Ma em`a iques, Vol 39 (1995), 233–240.
LIE SOLVABLE GROUP ALGEBRAS OF
DERIVED LENGTH THREE
Meena Sahai
Abs ac
Le Kbe a field o cha ac e is ic p>2 and le Gbe a g oup.
Necessa y and sufficien condi ions a e ob ained so ha he g oup
algeb a KG is s ongly Lie sol able o de i ed leng h a mos 3.
I is also shown ha hese condi ions a e equi alen o KG Lie
sol able o de i ed leng h 3 in cha ac e is ic p≥7.
1. In oduc ion
Any associa i e ing Rgi es ise o he associa ed Lie ing L(R) unde
he Lie mul iplica ion [x, y]=xy −yx,x,y∈R. We define, induc i ely,
[x1,x
2,... ,x
n]=[[x1,x
2,... ,x
n−1],x
n]. An addi i e subg oup Vo R
is called a Lie ideal o Ri [ , ]∈V o all ∈Vand ∈R. Fo any wo
Lie ideals Vand W, we deno e by [V,W] o be he addi i e subg oup o
Rgene a ed by {[ ,w]| ∈Vand w∈W}.
We define he Lie de i ed se ies δ[n](L(R)) and he s ong Lie de i ed
se ies δ(n)(R), n≥0, by induc ion as ollows:
δ[0](L(R)) = δ(0)(R)=R,
δ[n](L(R)) = [δ[n−1](L(R)),δ[n−1](L(R))],
δ[n](R)=[δ(n−1)(R),δ(n−1)(R)]R.
Ris Lie sol able o de i ed leng h ni δ[n](L(R)) = 0 bu δ[n−1](L(R)) =
0. Simila ly Ris s ongly Lie sol able o de i ed leng h ni δ(n)(R)=0
bu δ(n−1)(R)= 0. Lie sol able/s ongly Lie sol able ings o de i ed
leng h 2 a e called Lie me abelian/s ongly Lie me abelian ings. Also
Ris said o be Lie cen ally me abelian i [δ[2](L(R)),R]=0.
Le Kbe a field wi h Cha K=pand le Gbe a g oup. I is known
ha he g oup algeb a KG is Lie sol able i and only i Ghas a 2-abelian
subg oup o index a mos 2, when p= 2 and Gis p-abelian when p=2.
234 M. Sahai
Fo p= 2 his is equi alen o KG s ongly Lie sol able (see [4, Chap-
e V]). Bu he connec ion be ween he o de o he de i ed subg oup
Go Gand he de i ed leng h o KG is no known as ye . In his
di ec ion Le in and Rosenbe ge [1] ha e cha ac e ized Lie me abelian
g oup ings. Lie cen ally me abelian g oup algeb as ha e been s udied
by Sha ma and S i as a a [6] and Sahai and S i as a a [3]. I is shown
in [1] ha he g oup ing RG o a g oup Go e a commu a i e ing R
is Lie me abelian i and only i i is s ongly Lie me abelian.
In his pape , necessa y and sufficien condi ions o he g oup algeb a
KG, Cha K=p≥3, o be s ongly Lie sol able o de i ed leng h
a mos 3 ha e been ob ained. I is shown ha hese condi ions a e
equi alen o KG Lie sol able o de i ed leng h a mos 3 o p≥7.
2. Resul s and P oo s
Th oughou his sec ion Kdeno es a field wi h Cha K=p≥3 and
Gdeno es a g oup.
I is well known and easy o see ha i Mand Na e no mal subg oups
o G, hen
[∆(M)KG,∆(N)KG]KG = ∆((M,N))KG
+∆(M)∆(N)∆(G)KG
+ ∆((M,G))∆(N)KG
+∆(M)∆((N,G))KG.
In pa icula , i we ake M=N=G, hen
δ(2)(KG) = [∆(G)KG,∆(G)KG]KG
=∆(G)KG
+∆(G)3KG
+∆(γ3(G))∆(G)KG
+∆(G)∆(γ3(G))KG.
Rema k 2.1. I Gis cen al, hen he abo e equa ion gi es
δ(2)(KG)=∆(G)3KG. Fu he , i γ3(G)=G, hen δ(2)(KG)=
∆(G)2KG.
I |G|=pn, Cha K=pand (G) deno es he nilpo ency index o he
augmen a ion ideal ∆(G), hen i is known ha n(p−1)+1 ≤ (G)≤pn
wi h equali y on he le / igh hand side i and only i Gis elemen a y
abelian/cyclic (see [2]). This will be used o (G), as Gis a fini e
p-g oup i KG is Lie sol able. Fo any elemen x∈Gwe deno e ˆx=
1+x+x2+···+xn−1whe e o de o xis n.
We s a wi h he ollowing s aigh o wa d obse a ion.
Lie Sol able G oup Algeb as 235
Lemma 2.2. Fo all n≥1,∆(G)2n−1KG ⊆δ(n)(KG)⊆
∆(G)2n−1KG.
P oo : The igh hand side inclusion is immedia e by induc ion on n.
Since δ(1)(KG)=∆(G)KG and he iden i y
δ1δ2[g1,g
2]=[δ1g1,δ
2g2]−[δ1,δ
2g2]g1−[δ1g1,δ
2]g2+[δ1,δ
2]g1g2
is ue o all δ1,δ2∈∆(G)2n−1and g1,g2∈G,weha e
∆(G)2n+1−1KG =∆(G)2n−1∆(G)2n−1∆(G)KG
=∆(G)2n−1∆(G)2n−1[KG,KG]KG
⊆[∆(G)2n−1KG,∆(G)2n−1KG]KG
⊆[δ(n)(KG),δ(n)(KG)]KG
=δ(n+1)(KG).
This p o es he le hand side inclusion by induc ion on n.
Theo em 2.3. Le Kbe a field o cha ac e is ic p=2and le Gbe
a g oup. Then δ(3)(KG)=0i and only i one o he ollowing holds:
(i) Gis abelian.
(ii) p=7,G=C7and γ3(G)=1.
(iii) p=5,G=C5and ei he γ3(G)=1o γn(G)=G o all n≥3
wi h xg=x−1 o all x∈Gand o all g/∈CG(G).
(i ) p=3,Gis a g oup o one o he ollowing ypes:
(a) G=C3.
(b) G=C3×C3and ei he γ3(G)=1o γ3(G)=C3,γ4(G)=1
o γn(G)=G, o all n≥3wi h xg=x−1 o all x∈G
and o all g/∈CG(G).
(c) G=C3×C3×C3,γ3(G)=1.
P oo : Suppose ha δ(3)(KG) = 0. Since Cha K=2,Gis a fini e
p-g oup. Le |G|=pn. By Lemma 2.2, ∆(G)7KG ⊆δ(3)(KG)⊆
∆(G)4KG.Thus∆(G)7= 0. This in u n implies (G)≤7. By he
discussion ollowing jus a e Rema k 2.1, we conclude ha
(i) p≥11 implies n= 0. In his case G= 1 and hus Gis abelian.
(ii) p= 7 and Gis non-abelian implies n= 1 and G=C7.
(iii) p= 5 and Gis non-abelian implies n= 1 and G=C5.
(i ) p= 3 and Gis non-abelian implies Gis C3o C3×C3o C3×
C3×C3.
236 M. Sahai
I Gis abelian, we a e h ough. So we discuss each non-abelian case
sepa a ely.
Case (ii). p= 7. In his case G=C7. We shall show ha Gis
cen al, i.e., γ3(G) = 1. I no hen γ3(G)=Gand by Rema k 2.1,
δ(2)(KG)=∆(G)2KG. Le G=x. Then (x−1)6=ˆx. Now o any
g∈G,weha e
0=[(x−1)2,(x−1)2g−1]
=(x−1)3[x, g−1]+(x−1)2[x, g−1](x−1)
=−x{(x−1)3((x, g)−1)+(x−1)2((x, g)−1)(xg−1)}g−1.
I (x, g)=xk,1≤k≤5, hen we ge
(x−1)4(1 + x+x2+···+xk−1)(2 + x+x2+···+xk)=0.
Mul iplying by (x−1)2, we ha e k(k+2)ˆx=0. Thusk(k+2)=0in
K.Sok=5. Nowk= 5 is no possible because o he wise
(x−1)4(1 + x+x2+x3+x4)(2 + x+x2+x3+x4+x5)=0,
which gi es
(x−1)5(1 + x+x2+x3+x4)(5+4x+3x2+2x3+x4)=0.
Mul iplying by (x−1) we ge 75ˆx= 0 which is no ue. Thus k=0
and (x, g) = 1 o all g∈G. Hence Gis cen al.
Case (iii). p= 5. In his case G=C5. Le G=x.I Gis no
cen al, hen γ3(G)=G, and δ(2)(KG)=∆(G)2KG. Le (x, g)=xk
o some g∈G,1≤k≤3. I we p oceed exac ly as in he p e ious case,
we ge k(k+2)=0inK. This gi es k=3. Thus(x, g)=1o x3, i.e.,
i g/∈CG(G), hen xg=x−1, as desi ed.
Case (i ). p=3. I G=C3, we a e h ough. I G=C3×C3
hen (G) = 5 and γ3(G)=1o C3o G.I γ3(G) = 1, we a e
h ough. Conside he case when γ3(G)=C3=z. Le y∈Gsuch
ha y/∈γ3(G). Since ∆(γ3(G))∆(G)KG ⊆δ(2)(KG), we ha e ha o
all g∈G
0=[(z−1)(y−1)g−1,(z−1)(y−1)]
=(z−1)2(y−1)[g−1,y]+(z−1)(y−1)[g−1,z](y−1)
=(z−1)2(y−1)y(( y,g)−1)g−1+(z−1)(y−1)z(( z,g)−1)(yg−1)g−1
.
Lie Sol able G oup Algeb as 237
Fi s e m is ze o because (y,g)∈γ3(G) and ∆(γ3(G))3=0. Thus
(z−1)(y−1)(zk−1)(yg−1) = 0, whe e (z,g)=zk. This implies ha i
k= 0, hen (z−1)2(y−1)2= 0, because yg=y(y,g) and (y,g)∈γ3(G).
Bu his is a con adic ion o he ac ha y/∈γ3(G). Hence k= 0. This
shows ha γ3(G) is cen al, i.e., γ4(G)=1.
I γ3(G)=G, hen δ(2)(KG)=∆(G)2KG. Le G=x×yand
le g∈Gsuch ha g/∈CG(G). Now [(x−1)2,(x−1)2g−1] = 0. Using
he ac ha (x−1)3= 0 and expanding we ge (x−1)2(xg−1)2=0.
This implies xg∈x. Simila ly yg∈y. Suppose ha xg=x,so
xg=x−1.I yg=y, hen [(x−1)(y−1)g−1,(x−1)(y−1)] = 0 gi es
ha (x−1)2(y−1)2=0.Soy∈xwhich is no possible. Hence yg=y
and we mus ha e yg=y−1.Thusi g/∈CG(G), hen xg=x−1and
yg=y−1which p o es ha ug=u−1 o e e y u∈G.
Now i G=C3×C3×C3, hen (G) = 7. Also ∆(G)3(KG)⊆
δ(2)(KG) and ∆(γ3(G))∆(G)KG ⊆δ(2)(KG). We wish o p o e ha
Gis cen al, i.e., γ3(G) = 1. Suppose, i possible γ3(G)= 1. Le
1=x∈γ3(G). Choose y,z∈Gsuch ha G=x×y×z. As be o e
o any g∈G,[(x−1)2g−1,(x−1)2] = 0, implies (x−1)2(xg−1)2=
0. This implies xg∈x, i.e., xg=xo x−1. Nex obse e ha o
any u, ∈G, we ha e [u−1, ]={ u−1−( −1)}u−1.Weha e
[(y−1)(z−1)2g−1,(x−1)(z−1)] = 0 and so (y−1)(z−1)2({(xg−1) −
(x−1)}(zg−1) + (x−1){(zg−1) −(z−1)})=0. Le zg=x ysz .
I xg=x, we ge (y−1)(x−1)(x ys−1)ˆz= 0. This is possible only
i x ys= 1, because x,y,za e independen . I xg=x−1, we ge , a e
simplifica ion ha (y−1)(x−1−1)(x ys−1)ˆz= 0 and hence again
x ys=1. Thuszg∈z. Tha is, zg=zo z−1 o all g∈G. Fu he
o any g∈G,[(y−1)2(z−1)g−1,(x−1)(z−1)] = 0 implies
(y−1)2(z−1)({(xg−1)−(x−1)}(zg−1)+(x−1){(zg−1)−(z−1)})=0.
I zg=zand xg=x−1, hen we ge (y−1)2(z−1)2(x−1) = 0, which
is impossible since x,y,za e independen . Thus zg=zimplies xg=x.
I zg=z−1and xg=x, we again ge (y−1)2(z−1)2(x−1) = 0, which
is no possible as be o e. Finally i zg=z−1and xg=x−1, we ge
(y−1)2(z−1)2(x−1){x(z+1)+z}= 0, which is again no possible
because x,y,za e independen . Thus o any g∈G, we ha e zg=z
and xg=x. Simila ly o any g∈G, we mus ha e yg=y.ThusGis
cen al.
Now we p o e he con e se. I Gis abelian, hen clea ly δ(3)(KG)=0.
I Cha K= 7 and G=C7wi h γ3(G) = 1, hen by Rema k 2.1
δ(2)(KG)=∆(G)3KG.Thus
δ(3)(KG) = [∆(G)3KG,∆(G)3KG]KG
⊆∆(G)6[KG,KG]KG ⊆∆(G)7KG =0.

238 M. Sahai
I Cha K= 5 and G=C5=x, say, hen (G) = 5. Fi s le
γ3(G) = 1. Then as abo e δ(3)(KG)⊆∆(G)7KG = 0. Now le γ3(G)=
Gwi h he condi ion ha i g/∈CG(G), hen xg=x−1. Clea ly
∆(G)KG =(x−1)KG and by Rema k 2.1, δ(2)(KG)=∆(G)2KG.
Thus δ(3)(KG)=[(x−1)2KG,(x−1)2KG]KG. Le g1,g2∈G, hen
[(x−1)2g1,(x−1)2g2]
=(x−1)4[g1,g
2]+(x−1)3[g1,x]g2+(x−1)2[g1,x](x−1)g2
+(x−1)3[x, g2]g1+(x−1)2[x, g2](x−1)g1
=(x−1)4((g−1
1,g−1
2)−1)g2g1+(x−1)3((g−1
1,x
−1)−1)xg1g2
+(x−1)2((g−1
1,x
−1)−1)x(xg−1
1−1)g1g2
+(x−1)3((x−1,g−1
2)−1)xg−1
2g2g1
+(x−1)2((x−1,g−1
2)−1)xg−1
2(xg−1
2−1)g2g1
=(x−1)2((g−1
1,x
−1)−1)x(x−1)+(xg−1
1−1)g1g2
+(x−1)2((x−1,g−1
2)−1)xg−1
2(x−1)+(xg−1
2−1)g2g1
=(x−1)4((g−1
1,x
−1)−1)g1g2+(x−1)4((x−1,g−1
2)−1)g2g1
assuming ha xg1=x−1,xg2=x−1, o he cases gi e 0. The exp ession
on he igh hand side is 0 as (G) = 5. Thus δ(3)(KG)=0.
I Cha K= 3 and G=C3 hen (G) = 3 and so δ(3)(KG)⊆
∆(G)4KG =0.
Assume ha G=C3×C3. Then (G) = 5. Now i γ3(G) = 1 hen
δ(2)(KG)=∆(G)3KG and δ(3)(KG)⊆∆(G)7KG =0. I γ3(G)=C3
and γ4(G) = 1, hen
δ(2)(KG)=∆(G)3KG +∆(γ3(G))∆(G)KG.
δ(3)(KG) = 0 because (G)=5and (γ3(G))=3. I γ3(G)=Gand
ug=u−1 o e e y u∈G,g/∈CG(G), assume ha G=x×y.
Then ∆(G)KG =(x−1)KG +(y−1)KG and ∆(G)2KG =(x−
1)2KG+(y−1)2KG+(x−1)(y−1)KG. Also δ(2)(KG)=∆(G)2KG.
Fo any g/∈CG(G) and u, ∈G, we ha e
[g,(u−1)( −1)]=(u−1)[g, ]+[g,u]( −1)
=(u−1)((g−1, −1)−1) g+((g−1,u
−1)−1)ug( −1)
=(u−1)( −1) g +(u−1)u( g−1−1)g
=(u−1)( −1)( +u +u)g∈∆(G)3KG.
Lie Sol able G oup Algeb as 239
Using his and abo e we see ha δ(3)(KG)⊆∆(G)5KG =0.
The case when G=C3×C3×C3and γ3(G) = 1 ollows easily as
(G)=7andδ(2)(KG)=∆(G)3KG.
Example 2.4. By abo e Theo em δ(3)(KD10) = 0, i Cha K=5
whe e D10 deno es he Dihed al g oup o o de 10. Thus in his case
G=C5need no be cen al and Gneed no e en be nilpo en . Also i G
is he semidi ec p oduc o C3×C3by C2induced by he au omo phism
sending e e y elemen o C3×C3 o i s in e se, hen δ(3)(KG) = 0 whe e
Cha K= 3. Thus in Cha K= 3 also Gneed no be nilpo en .
Co olla y 2.5. Le Kbe a field wi h Cha K=p≥7and le Gbe a
g oup. Then he ollowing a e equi alen :
(i) δ(3)(KG)=0,
(ii) δ[3](L(KG)) = 0.
P oo : Clea ly i δ(3)(KG) = 0, hen δ[3](L(KG)) = 0. Sup-
pose ha δ[3](L(KG)) = 0. Le x,y∈G. Then by [6,
Lemma 2.4(iii)] 2((x, y, y)−1)3∈γ3(δ[1](L(KG))). Since by [5,
Lemma 1.7] [γ3(δ[1](L(KG)))]2KG ⊆δ[3](L(KG))KG, we ge ha
4((x, y, y)−1)6= 0. Since Gis a p-g oup and p≥7, (x, y, y)=1and
Gis 2-Engel. I is well known ha o a 2-Engel g oup G,(G,G)3=1.
Again because Gis a p-g oup, p≥7, we conclude ha γ3(G) = 1. Now
[[x, y][x, y, y],[x, y]]=[[x, y],[xy, y],[x, y]] is in γ3(δ[1](L(KG))). The e-
o e as abo e
0=[[x, y][x, y, y],[x, y]]2
=[yx((x, y)−1)[yx((x, y)−1),y],yx((x, y)−1)]2
=[yxy[x, y],yx]2((x, y)−1)6
=[yxy2x((x, y)−1),yx]2((x, y)−1)6
=[yxy2x, yx]2((x, y)−1)8
=(yxy2xyx)2((x, y)−1)10.
This gi es ha ((x, y)−1)10 = 0. Bu Gis a p-g oup and hence o
p≥11, (x, y) = 1 and o p=7,(x, y)7= 1. We conclude ha o
p≥11, Gis abelian and o p=7,G7= 1. Now le p= 7 and le x,y,
u, ∈G, hen
[[x, y][x, y, y],[x, y]][[u, ][u, , ],[u, ]] ∈[γ3(δ[1](L(KG)))]2
⊆δ[3](L(KG)) = 0.
240 M. Sahai
Simpli ying as abo e we ge ((x, y)−1)5((u, )−1)5= 0. And so (u, )∈
(x, y).ThusGis cyclic. Res ollows om Theo em 2.3.
Nex we gi e an example o illus a e ha he e a e g oup algeb as
which a e Lie sol able o leng h h ee bu no s ongly Lie sol able o
leng h h ee.
Example 2.6. Le K=Z2and G=S3, he symme ic g oup on
h ee le e s. Then Gis a cyclic g oup o o de h ee. I can be easily
e ified ha δ[2](L(KG)) ⊆KGand hence δ[3](L(KG)) = 0. Bu Z2S3
is no s ongly Lie sol able.
Acknowledgemen s. The au ho is hank ul o he e e ee o alu-
able sugges ions.
Re e ences
1. F. Le in and G. Rosenbe ge , Lie me abelian g oup ings,
P ep in no. 60, Ruh -Uni e si ¨a , Bochum, Dec. 1985.
2. K. Mo ose and Y. Ninomiya, On he nilpo ency index o he
adical o a g oup algeb a, Hokkaido Ma h. J. 4(1975), 261–264.
3. M. Sahai and J. B. S i as a a, A no e on Lie cen ally
me abelian g oup algeb as, o appea in J. Algeb a.
4. S. K. Sehgal,“Topics in G oup Rings,” Ma cel Dekke , 1978.
5. R. K. Sha ma and J. B. S i as a a, Lie sol able ings, P oc.
Ame . Ma h. Soc. 94 (1985), 1–8.
6. R. K. Sha ma and J. B. S i as a a, Lie cen ally me abelian
g oup ings, J. Algeb a 151(2) (1992), 476–486.
Depa men o Ma hema ics
Indian Ins i u e o Technology
New Delhi - 110 016
INDIA
P ime a e si´o ebuda el 6 de Juliol de 1994,
da e a e si´o ebuda el 29 de Juny de 1995