let G be Group; for G1 being Subgroup of G
for N being strict normal Subgroup of G st N is Subgroup of G1 & G1 ./. ((G1,N) `*`) is Subgroup of center (G ./. N) holds
[.G1,((Omega). G).] is Subgroup of N
let G1 be Subgroup of G; for N being strict normal Subgroup of G st N is Subgroup of G1 & G1 ./. ((G1,N) `*`) is Subgroup of center (G ./. N) holds
[.G1,((Omega). G).] is Subgroup of N
let N be strict normal Subgroup of G; ( N is Subgroup of G1 & G1 ./. ((G1,N) `*`) is Subgroup of center (G ./. N) implies [.G1,((Omega). G).] is Subgroup of N )
assume that
A1:
N is Subgroup of G1
and
A2:
G1 ./. ((G1,N) `*`) is Subgroup of center (G ./. N)
; [.G1,((Omega). G).] is Subgroup of N
A3:
(G1,N) `*` = N
by A1, GROUP_6:def 1;
reconsider J = G1 ./. ((G1,N) `*`) as Subgroup of G ./. N by A1, GROUP_6:28;
reconsider I = N as normal Subgroup of G1 by A3;
A4:
commutators (G1,((Omega). G)) c= carr N
proof
now for x being Element of G st x in commutators (G1,((Omega). G)) holds
x in carr Nlet x be
Element of
G;
( x in commutators (G1,((Omega). G)) implies x in carr N )assume
x in commutators (
G1,
((Omega). G))
;
x in carr Nthen consider a,
b being
Element of
G such that A5:
(
x = [.a,b.] &
a in G1 &
b in (Omega). G )
by GROUP_5:52;
reconsider c =
a as
Element of
G1 by A5, STRUCT_0:def 5;
reconsider S9 =
c * I as
Element of
J by A3, GROUP_6:22;
A6:
S9 in J
by STRUCT_0:def 5;
reconsider T =
b * N as
Element of
(G ./. N) by GROUP_6:14;
reconsider d =
c as
Element of
G ;
d * N = c * I
by GROUP_6:2;
then reconsider S =
S9 as
Element of
(G ./. N) by GROUP_6:14;
S in center (G ./. N)
by A2, A6, GROUP_2:40;
then A7:
S * T = T * S
by GROUP_5:77;
A8:
(
S = d * N &
T = b * N &
@ S = S &
@ T = T )
by GROUP_6:2;
then A9:
S * T =
(d * N) * (b * N)
by GROUP_6:def 3
.=
d * (N * (b * N))
by GROUP_3:10
.=
d * ((N * b) * N)
by GROUP_3:13
.=
d * ((b * N) * N)
by GROUP_3:117
.=
d * (b * N)
by GROUP_6:5
.=
(d * b) * N
by GROUP_2:105
;
T * S =
(b * N) * (d * N)
by A8, GROUP_6:def 3
.=
b * (N * (d * N))
by GROUP_3:10
.=
b * ((N * d) * N)
by GROUP_3:13
.=
b * ((d * N) * N)
by GROUP_3:117
.=
b * (d * N)
by GROUP_6:5
.=
(b * d) * N
by GROUP_2:105
;
then A10:
((d ") * (b ")) * ((d * b) * N) =
(((d ") * (b ")) * (b * d)) * N
by A7, A9, GROUP_2:105
.=
((d ") * ((b ") * (b * d))) * N
by GROUP_1:def 3
.=
((d ") * (((b ") * b) * d)) * N
by GROUP_1:def 3
.=
((d ") * ((1_ G) * d)) * N
by GROUP_1:def 5
.=
((d ") * d) * N
by GROUP_1:def 4
.=
(1_ G) * N
by GROUP_1:def 5
.=
carr N
by GROUP_2:109
;
((d ") * (b ")) * ((d * b) * N) =
(((d ") * (b ")) * (d * b)) * N
by GROUP_2:105
.=
[.d,b.] * N
by GROUP_5:16
;
then
[.d,b.] in N
by A10, GROUP_2:113;
hence
x in carr N
by A5, STRUCT_0:def 5;
verum end;
hence
commutators (
G1,
((Omega). G))
c= carr N
;
verum
end;
gr (carr N) = N
by GROUP_4:31;
hence
[.G1,((Omega). G).] is Subgroup of N
by A4, GROUP_4:32; verum