3.2 The Group Structure
25
has very remarkable properties. Each row and each column represent a permutation
of the ordered set of elements, but in such a way that every element occurs only
once in each row and column. This is a direct consequence of the group properties.
As in many group-theoretical proofs, the simplest way to show this is by a reductio
ad absurdum. Suppose that a given element, ˆ
T , occurred at entries ij and ik, with
ˆ
R k = ˆ
R j . Then one would have, by applying the rules:
ˆ
R i ˆ
R j = ˆ
R i ˆ
R k
ˆ
R
−1
i ( ˆ
R i ˆ
R j ) = ˆ
R
−1
i ( ˆ
R i ˆ
R k )
ˆ
R
−1
i
ˆ
R i
ˆ
R j =
ˆ
R
−1
i
ˆ
R i
ˆ
R k
ˆ
E ˆ
R j = ˆ
E ˆ
R k
ˆ
R j = ˆ
R k
(3.9)
which would contradict the original supposition. Along the same lines it is easy to
prove that the inverse of a product is equal to the product of the inverses in the
opposite order:
( ˆ
R i ˆ
R j )
−1 = ˆ
R
−1
j
ˆ
R
−1
i
(3.10)
As a matter of principle, the group multiplication table contains everything there is
to know about the group. It is, though, not necessary to store the whole multiplication table. A more compact way uses generators. The generators are defined as a
minimal set of elements capable of generating the whole group. For the present example, two generators are needed, e.g., ˆ
C 3 and ˆ
σ 1 . It is sufficient to make all binary
combinations of these two operators in order to generate all remaining elements:
ˆ
C 3 ˆ
C 3 = ˆ
C
2
3
ˆ
σ 1 ˆ
σ 1 = ˆ
E
ˆ
C 3 ˆ
σ 1 =ˆ σ 3
ˆ
σ 1 ˆ
C 3 =ˆ σ 2
(3.11)
Alternatively, any pair of reflection planes would suffice as generators, say ˆ
σ 1
and ˆ
σ 2 , but in this case the remaining symmetry plane can be obtained only by a
further multiplication:
ˆ
σ 1 ˆ
σ 1 = ˆ
E
ˆ
σ 2 ˆ
σ 2 = ˆ
E
ˆ
σ 1 ˆ
σ 2 = ˆ
C 3
ˆ
σ 2 ˆ
σ 1 = ˆ
C
2
3
ˆ
σ 1 ˆ
σ 2 ˆ
σ 1 =ˆ σ 3
(3.12)
25
has very remarkable properties. Each row and each column represent a permutation
of the ordered set of elements, but in such a way that every element occurs only
once in each row and column. This is a direct consequence of the group properties.
As in many group-theoretical proofs, the simplest way to show this is by a reductio
ad absurdum. Suppose that a given element, ˆ
T , occurred at entries ij and ik, with
ˆ
R k = ˆ
R j . Then one would have, by applying the rules:
ˆ
R i ˆ
R j = ˆ
R i ˆ
R k
ˆ
R
−1
i ( ˆ
R i ˆ
R j ) = ˆ
R
−1
i ( ˆ
R i ˆ
R k )
ˆ
R
−1
i
ˆ
R i
ˆ
R j =
ˆ
R
−1
i
ˆ
R i
ˆ
R k
ˆ
E ˆ
R j = ˆ
E ˆ
R k
ˆ
R j = ˆ
R k
(3.9)
which would contradict the original supposition. Along the same lines it is easy to
prove that the inverse of a product is equal to the product of the inverses in the
opposite order:
( ˆ
R i ˆ
R j )
−1 = ˆ
R
−1
j
ˆ
R
−1
i
(3.10)
As a matter of principle, the group multiplication table contains everything there is
to know about the group. It is, though, not necessary to store the whole multiplication table. A more compact way uses generators. The generators are defined as a
minimal set of elements capable of generating the whole group. For the present example, two generators are needed, e.g., ˆ
C 3 and ˆ
σ 1 . It is sufficient to make all binary
combinations of these two operators in order to generate all remaining elements:
ˆ
C 3 ˆ
C 3 = ˆ
C
2
3
ˆ
σ 1 ˆ
σ 1 = ˆ
E
ˆ
C 3 ˆ
σ 1 =ˆ σ 3
ˆ
σ 1 ˆ
C 3 =ˆ σ 2
(3.11)
Alternatively, any pair of reflection planes would suffice as generators, say ˆ
σ 1
and ˆ
σ 2 , but in this case the remaining symmetry plane can be obtained only by a
further multiplication:
ˆ
σ 1 ˆ
σ 1 = ˆ
E
ˆ
σ 2 ˆ
σ 2 = ˆ
E
ˆ
σ 1 ˆ
σ 2 = ˆ
C 3
ˆ
σ 2 ˆ
σ 1 = ˆ
C
2
3
ˆ
σ 1 ˆ
σ 2 ˆ
σ 1 =ˆ σ 3
(3.12)