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Basis for Octonions:
    To construct basal elements of octonion from that of quaternions, let
    ir  ( r = 0,1,2,3 ) denote quaternion basis, and
    em  ( m = 0,1,...,7 ) denote octonion basis.
Let    em  :=  ( im ; 0 ) ,     for m  =  0,1,...,3,      and
         em  :=  ( 0 ; im-4 )  ,  for m  =  4,...,7.
e.g.    e0  =  ( i0 ; 0 ) = ( 1 ; 0 ) = 1,
          e2 = ( i2  ; 0 ) ,            e7 = ( 0 ; i3 ) .
Then the multiplication rule for the octonion basal elements can be found using Cayley-Dickson process (when  m is chosen as -1 at each iteration).   For example,
        e4 e5 = ( 0 ; 1 ) ( 0 ; i1 ) = ( - i1* ; 0 ) = ( i1 ; 0 ) = e1  .

    A multiplication table can then be constructed:
 
 
e0 = 1
e1
e2
 e3
e4
e5
e6
e7
e0 = 1
1
 e1
 e2
 e3
 e4
 e5
 e6
 e7
e1
 e1
-1
e3
 -e2
 e5
 -e4
 -e7
 e6
e2
 e2
 -e3
-1
 e1
 e6
 e7
 -e4
 -e5
e3
 e3
 e2
 -e1
-1
 e7
 -e6
 e5
 -e4
e4
 e4
 -e5
 -e6
 -e7
-1
 e1
 e2
 e3
e5
 e5
 e4
 -e7
 e6
 -e1
-1
 -e3
 e2
e6
 e6
 e7
 e4
 -e5
 -e2
 e3
-1
-e1
e7
 e7
 -e6
 e5
 e4
 -e3
 -e2
 e1
 -1

    The multiplication rule for the pure octonion basis ( ej , j = 1,2,...,7 ) can be represented diagrammatically by
 

where the arrows indicate the signs of the resulting basal elements.
e.g.,    e2 i7  =  - e5 ,    e3 e4  =  e7  .

    Another more text friendly way to express the multiplication rule is by means of octonion cycles.   The multiplication of 2 octonion basal elements, ej   and ek , is given by
     ej   ek    =  -  dj k  + (jkm)S e jkm e m  ,    ( j, k, m = 1,2,...,7 )   ,
where the summation is over all possible permutation of (j,k,m), and the structure constant  ejkm is totally anti-symmetric in its indices whose value is determined from the following octonion cycles:
        ( 1,2,3 ),  ( 1,4,5 ),  ( 1,7,6 ),  ( 2,4,6 ),  ( 2,5,7 ),  ( 3,4,7 ),  ( 3,6,5 ) ,

and is given by
 
e jkm   = {   1   if  (j,k,m) is an even permutation,
- 1   if  (j,k,m) is an odd permutation,
  0   otherwise,
 of an octonion cycle,

and as usual
 
 
dj k   1   if j = k
0   otherwise.

.