let G be Group; for N1, N2 being strict normal Subgroup of G ex M being strict Subgroup of G st the carrier of M = N1 * N2
let N1, N2 be strict normal Subgroup of G; ex M being strict Subgroup of G st the carrier of M = N1 * N2
A1:
1_ G in N1 * N2
A3:
for x, y being Element of G st x in N1 * N2 & y in N1 * N2 holds
x * y in N1 * N2
proof
let x,
y be
Element of
G;
( x in N1 * N2 & y in N1 * N2 implies x * y in N1 * N2 )
assume that A4:
x in N1 * N2
and A5:
y in N1 * N2
;
x * y in N1 * N2
consider a,
b being
Element of
G such that A6:
(
x = a * b &
a in N1 &
b in N2 )
by A4, Th6;
consider c,
d being
Element of
G such that A7:
(
y = c * d &
c in N1 &
d in N2 )
by A5, Th6;
A8:
x * y =
((a * b) * c) * d
by A6, A7, GROUP_1:def 3
.=
(a * (b * c)) * d
by GROUP_1:def 3
;
b * c in N2 * N1
by A6, A7, Th6;
then
b * c in N1 * N2
by GROUP_3:125;
then consider b9,
c9 being
Element of
G such that A9:
(
b * c = b9 * c9 &
b9 in N1 &
c9 in N2 )
by Th6;
A10:
x * y =
((a * b9) * c9) * d
by A8, A9, GROUP_1:def 3
.=
(a * b9) * (c9 * d)
by GROUP_1:def 3
;
A11:
a * b9 in N1
by A6, A9, GROUP_2:50;
c9 * d in N2
by A7, A9, GROUP_2:50;
hence
x * y in N1 * N2
by A10, A11, Th6;
verum
end;
for x being Element of G st x in N1 * N2 holds
x " in N1 * N2
hence
ex M being strict Subgroup of G st the carrier of M = N1 * N2
by A1, A3, GROUP_2:52; verum