let k, n be Element of NAT ; for G being finite strict Group
for a being Element of G st G = gr {a} & card G = n holds
( G = gr {(a |^ k)} iff k gcd n = 1 )
let G be finite strict Group; for a being Element of G st G = gr {a} & card G = n holds
( G = gr {(a |^ k)} iff k gcd n = 1 )
let a be Element of G; ( G = gr {a} & card G = n implies ( G = gr {(a |^ k)} iff k gcd n = 1 ) )
assume that
A1:
G = gr {a}
and
A2:
card G = n
; ( G = gr {(a |^ k)} iff k gcd n = 1 )
A3:
( G = gr {(a |^ k)} implies k gcd n = 1 )
proof
set d =
k gcd n;
assume that A4:
G = gr {(a |^ k)}
and A5:
k gcd n <> 1
;
contradiction
k gcd n divides n
by NAT_D:def 5;
then consider p being
Nat such that A6:
n = (k gcd n) * p
by NAT_D:def 3;
A7:
p <> 0
by A2, A6;
A8:
p <> n
k gcd n divides k
by NAT_D:def 5;
then consider l being
Nat such that A9:
k = (k gcd n) * l
by NAT_D:def 3;
A10:
not
a |^ k is
being_of_order_0
by GR_CY_1:6;
(a |^ k) |^ p =
a |^ ((l * (k gcd n)) * p)
by A9, GROUP_1:35
.=
a |^ (n * l)
by A6
.=
(a |^ n) |^ l
by GROUP_1:35
.=
(1_ G) |^ l
by A2, GR_CY_1:9
.=
1_ G
by GROUP_1:31
;
then A11:
ord (a |^ k) <= p
by A7, A10, GROUP_1:def 11;
p divides n
by A6, NAT_D:def 3;
then
p <= n
by A2, NAT_D:7;
then
p < n
by A8, XXREAL_0:1;
hence
contradiction
by A2, A4, A11, GR_CY_1:7;
verum
end;
( k gcd n = 1 implies G = gr {(a |^ k)} )
hence
( G = gr {(a |^ k)} iff k gcd n = 1 )
by A3; verum