E1.  If a and b commute , then ord(ab) is a devisor of lcm( m,n )

       Proof

Let lcm[m,n] = mx = ny   for some integer x,y
(ab) lcm[m,n] = a lcm[m,n]. b lcm[m,n]
= amx. bny = ex.ey = e
    o(ab) | lcm[m,n]

E2.  If m and n relatively prime , then no power of a can be equal to any power of b (except for e). ( Remark : Two integer are said to be relatively prime if they have no common factors except ? 1.)

       Proof

Let     0 < i < m และ   0 < j < n
suppose    ai = bj   some  i,j
e = am =  ei = (am)i = ami = (ai)m = (bj)m =bjm
so    n | mj   but   gcd(m,n) = 1
   n | j
It's conflict because   j<n
    ai bj    for all integer    i,j which not 0

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E3.  If m and n are relatively prime , then the products aⁱbⁱ ( 0 i < m , 0 j < n ) are all distinct.

       Proof

suppose    aibj  =  akbl ( 0 < i,k < m   and  0 < j,l < n )
ai - k   =  bl - j
from   E2  hence   i k = 0  and  l j = 0
so   i = k   and   l = j
    aibj  are  all  distinct.

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E4.  Let a and b commute. If m and n are relatively prime , then ord(ab) = mn.

       Proof

from part  E1   hence  o(ab) | lcm[m,n]
from   mn = lcm[m,n].gcd(m,n) = lcm[m,n]
e  = (ab)^o(ab)  =  a ^lcm[m,n]. b ^lcm[m,n] ..(*)
(*)  square  m; e^m  =  (a ^lcm[m,n]. b ^lcm[m,n])^m  =  (a ^lcm[m,n])^m(b ^lcm[m,n])^m
= a ^m.lcm[m,n] . b ^m.lcm[m,n]  =  e^lcm[m,n] .b^m.lcm[m,n]
e  =  b ^m.lcm[m,n]
hence   n | m.lcm[m,n]   but  gcd(m,n) = 1
so   n | lcm[m,n]
in the same way
(*) square  n ; eⁿ  =  (a ^lcm[m,n]. b ^lcm[m,n])ⁿ  =  a ^n.lcm[m,n]
    m | n.lcm[m,n]   but  gcd(m,n) = 1
   m | lcm[m,n]
from    (1) m | lcm[m,n]   and   n | lcm[m,n]
(2) If   m|d   and   n|d   some integer  d   and  lcm[m,n] | d
and from above proofment  :  m | o(ab) , n | o(ab)  and  o(ab) | lcm[m,n]
and  from  lcm[m,n] | o(ab)
   o(ab) = lcm[m,n] = mn

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E5.  Let a and b commute. There is an element c in G whose order is lcm[m.n].

       Proof

Let     gcd(m,n) = q
hence      m  =  mq , n = nq
fix  c =  aⁱb  and  i = gcd(m,n)
from gcd(m,n) = q
  can write  q  in   q = mx + ny   form  some integer x,y
q = mx + ny
1 = (m/q)x + (n/q)y
1 = mx + ny
   gcd(m,n) = 1
    o(c) = o(aⁱb) = o(ab) = mn = lcm[m,n]

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E6.   Give an example to show that part 1 is not true if a and do not commute.

       Proof

Select  Permutation group : S₃
because it have not commutation
let a = (1 3) , b = (2 3)
hence   o(a) = 2   and   o(b) = 2 lcm[2,2] = 2
ab = (1 3)(2 3) = (1 2 3)
ba = (2 3)(1 3) = (2 1 3)
which   o(ab) = 3 
that means  o(ab) | lcm[m,n]   not true

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