D1.   If a^p = e where p is a prime number , then a has order p. ( a e.)

       Proof

from            a^p  =  e
     o(a) | p
but  p  is a prime number and 0 < |G|
hence  o(a)  =  p (because a e)

D2.   The order of a^k is a divisor ( factor ) of the order of a.

       Proof

from             a^o(a)    =     e
(a^o(a))^k    =     e^k      =     e
a^k.o(a)      =    (a^k) ^o(a)     =      e
     o(a^k) | o(a)

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D3.  If ord( a ) = km , then ord( a^k ) = m.

       Proof

e   =    a^km   =    (a^k)^m
  o(a^k) | m
want to show that   m | o(a^k)
let    x   =   o(a^k)
(a^k)^x   =   a^kx  =   e   o(a) | kx
from o(a)  =  km  and  x  =  o(a^k)
   km | k.o(a^k)
m | o(a^k)   o(a^k)  =  m

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D4.  If ord(a ) = n where n is odd , then ord(a² ) = n.

       Proof

hence     aⁿ  =  e
(aⁿ)²  =  e²  =  e  =  a²ⁿ   =  (a²)ⁿ o(a²) | n
let   m  =  o(a²)
so      (a²)^m   =   e   =   a^2m
   n | 2m
but    gcd(n,2)  =  1 (because n is odd)
hence   n | m         o(a²) = n

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D5.  If a has order n , and a^r = a^s , then n is a factor of r s.

       Proof

a^r    =   a^s
a^{r - s}   =    e
hence   o(a) | r-s
    n | r-s

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D6.  If a is the only element of order k in G , then a is in the center of G.

       Proof

from theorem C4
hence   k = o(a) = o(xax^-1)
a = xax^-1   ( because we have only a is the  element of order k )
ax = xa
     a    C

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D7.  If the order of a is not a multiple of m , then the order of a^k is not a multiple of m.

       Proof

proof  with Contrapositive form
hence    o(a^k)  =  mx บาง  x Z
want to show that    o(a)  =  my  some  y Z
from   D2   hence  o(a^k) | o(a)
o(a)  =  o(a^k).n    some    n Z
=  (mx)n   =   m(xn)  =  my ( x,n Z    xn Z ,  xn = y)
    can write o(a)   in  mutiple of  m form

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D8.  If ord( a ) = mk and a^rk = e , then r is a multiple of m.

       Proof

from    a^rk  =  e    o(a) | rk
hence   rk = [o(a)]x   some   x Z
and   o(a) = km
   rk = (km)x
r = mx
   can write   r in multiple of  m  form

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