let Gc be finite strict cyclic Group; ( ex k being Element of NAT st card Gc = 2 * k implies ex g1 being Element of Gc st
( ord g1 = 2 & ( for g2 being Element of Gc st ord g2 = 2 holds
g1 = g2 ) ) )
set n = card Gc;
given k being Element of NAT such that A1:
card Gc = 2 * k
; ex g1 being Element of Gc st
( ord g1 = 2 & ( for g2 being Element of Gc st ord g2 = 2 holds
g1 = g2 ) )
consider g being Element of Gc such that
A2:
Gc = gr {g}
by GR_CY_1:def 7;
A3:
ord g = card Gc
by A2, GR_CY_1:7;
take
g |^ k
; ( ord (g |^ k) = 2 & ( for g2 being Element of Gc st ord g2 = 2 holds
g |^ k = g2 ) )
A4: (g |^ k) |^ 2 =
g |^ (card Gc)
by A1, GROUP_1:35
.=
1_ Gc
by GR_CY_1:9
;
A5:
k <> 0
by A1;
A6:
for p being Nat st (g |^ k) |^ p = 1_ Gc & p <> 0 holds
2 <= p
not g |^ k is being_of_order_0
by GR_CY_1:6;
hence
ord (g |^ k) = 2
by A4, A6, GROUP_1:def 11; for g2 being Element of Gc st ord g2 = 2 holds
g |^ k = g2
let g2 be Element of Gc; ( ord g2 = 2 implies g |^ k = g2 )
consider k1 being Element of NAT such that
A10:
g2 = g |^ k1
by A2, Th6;
assume that
A11:
ord g2 = 2
and
A12:
g |^ k <> g2
; contradiction
now contradictionA13:
not
g is
being_of_order_0
by GR_CY_1:6;
consider t,
t1 being
Nat such that A14:
k1 = (k * t) + t1
and A15:
t1 < k
by A5, NAT_1:17;
A16:
2
* t1 < card Gc
by A1, A15, XREAL_1:68;
t1 <> 0
then A18:
2
* t1 <> 0
;
1_ Gc =
(g |^ k1) |^ 2
by A11, A10, GROUP_1:41
.=
g |^ (((k * t) + t1) * 2)
by A14, GROUP_1:35
.=
g |^ (((card Gc) * t) + (t1 * 2))
by A1
.=
(g |^ ((card Gc) * t)) * (g |^ (t1 * 2))
by GROUP_1:33
.=
((g |^ (ord g)) |^ t) * (g |^ (t1 * 2))
by A3, GROUP_1:35
.=
((1_ Gc) |^ t) * (g |^ (t1 * 2))
by GROUP_1:41
.=
(1_ Gc) * (g |^ (t1 * 2))
by GROUP_1:31
.=
g |^ (2 * t1)
by GROUP_1:def 4
;
hence
contradiction
by A3, A18, A16, A13, GROUP_1:def 11;
verum end;
hence
contradiction
; verum