let G be Group; for A being Subgroup of G holds
( A is normal Subgroup of G iff [.A,((Omega). G).] is Subgroup of A )
let A be Subgroup of G; ( A is normal Subgroup of G iff [.A,((Omega). G).] is Subgroup of A )
thus
( A is normal Subgroup of G implies [.A,((Omega). G).] is Subgroup of A )
by Lm3; ( [.A,((Omega). G).] is Subgroup of A implies A is normal Subgroup of G )
assume A1:
[.A,((Omega). G).] is Subgroup of A
; A is normal Subgroup of G
for b being Element of G holds b * A c= A * b
proof
let b be
Element of
G;
b * A c= A * blet x be
set ;
TARSKI:def 3 ( not x in b * A or x in A * b )
assume A2:
x in b * A
;
x in A * b
then reconsider x =
x as
Element of
G ;
consider Z being
Element of
G such that A3:
(
x = b * Z &
Z in A )
by A2, GROUP_2:125;
A4:
Z " in A
by A3, GROUP_2:60;
b " in (Omega). G
by STRUCT_0:def 5;
then A5:
[.(b " ),(Z " ).] in [.((Omega). G),A.]
by A4, GROUP_5:74;
[.(b " ),(Z " ).] in A
by A1, A5, GROUP_2:49;
then A6:
(((b * Z) * (b " )) * (Z " )) * Z in A
by A3, GROUP_2:59;
A7:
(((b * Z) * (b " )) * (Z " )) * Z =
((b * Z) * (b " )) * ((Z " ) * Z)
by GROUP_1:def 4
.=
((b * Z) * (b " )) * (1_ G)
by GROUP_1:def 6
.=
(b * Z) * (b " )
by GROUP_1:def 5
;
((b * Z) * (b " )) * b =
(b * Z) * ((b " ) * b)
by GROUP_1:def 4
.=
(b * Z) * (1_ G)
by GROUP_1:def 6
.=
b * Z
by GROUP_1:def 5
;
hence
x in A * b
by A3, A6, A7, GROUP_2:126;
verum
end;
hence
A is normal Subgroup of G
by GROUP_3:141; verum