3.6 Classes
33
The proof makes use of the coset concept. We start by considering all elements
that stabilize a given element, say ˆ
A 0 , of the group. One can prove that these elements constitute a subgroup H ⊂ G. Hence, we write:
ˆ
h x ∈ H ⇔ ˆ
h x ˆ
A 0 ˆ
h
−1
x = ˆ
A 0
(3.26)
The proof consists of a check of the four group criteria. The closure, for instance, is
proven as follows: suppose that both ˆ
h x and ˆ
h y stabilize ˆ
A 0 ; then their product will
also be a stabilizer:
ˆ
h x ˆ
h y ˆ
A 0 ( ˆ
h x ˆ
h y )
−1 = ˆ
h x ˆ
h y ˆ
A 0 ˆ
h
−1
y
ˆ
h
−1
x = ˆ
h x ˆ
A 0 ˆ
h
−1
x = ˆ
A 0
(3.27)
where we use the result of Eq. (3.10) that the inverse of a product is equal to the
product of the inverses in the reverse order. Next, we expand G in cosets of this
newly found subgroup H . All elements of a coset ˆ
R i H , with ˆ
R i /
∈ H , will transform
ˆ
A 0 into the same new element ˆ
A i :
ˆ
R i ˆ
h x ˆ
A 0 ( ˆ
R i ˆ
h x )
−1 = ˆ
R i ˆ
h x ˆ
A 0 ˆ
h
−1
x
ˆ
R
−1
i = ˆ
R i ˆ
A 0 ˆ
R
−1
i = ˆ
A i
(3.28)
The result must be different from ˆ
A 0 because, otherwise, ˆ
R i w o u l db ea ne l e m e n t
of H . In order to prove the theorem, the following remaining questions have to be
decided. Do different cosets give rise to different similarity transforms? Does one
obtain all elements of a class by finding all transforms of a given starting element?
The answers to both questions are affirmative. For the first question, if ˆ
R i and ˆ
R j
represent different cosets, one should conclude that ˆ
A i = ˆ
A j . Suppose that the opposite is true:
ˆ
R i ˆ
A 0 ˆ
R
−1
i = ˆ
R j ˆ
A 0 ˆ
R
−1
j
ˆ
R
−1
i
ˆ
R j ˆ
A 0 ˆ
R
−1
j
ˆ
R i = ˆ
A 0
ˆ
R
−1
i
ˆ
R j ˆ
A 0
ˆ
R
−1
i
ˆ
R j
−1 = ˆ
A 0
(3.29)
This implies that ˆ
R
−1
i
ˆ
R j stabilizes ˆ
A 0 and thus must belong to H , where it corresponds to, say, ˆ
h z . But one then again has
ˆ
R i ˆ
h z = ˆ
R i ˆ
R
−1
i
ˆ
R j = ˆ
R j
(3.30)
and thus ˆ
R j is a representative of the same coset as ˆ
R i , which contradicts the starting
assumption. Hence, there will be at least as many equivalent elements in the class as
there are cosets of the stabilizing subgroup. Have we then generated the entire class?
Yes, because by going through all the cosets, we run through the entire group. In this
way, we have found all elements that are conjugate to a given one, but, because of
transitivity, this also means that there cannot be other conjugate elements. The oneto-one mapping between conjugate elements and cosets implies that the number in
a class is equal to the number of cosets of the stabilizing subgroup and hence—by
Lagrange’s theorem—must be a divisor of the group order.
33
The proof makes use of the coset concept. We start by considering all elements
that stabilize a given element, say ˆ
A 0 , of the group. One can prove that these elements constitute a subgroup H ⊂ G. Hence, we write:
ˆ
h x ∈ H ⇔ ˆ
h x ˆ
A 0 ˆ
h
−1
x = ˆ
A 0
(3.26)
The proof consists of a check of the four group criteria. The closure, for instance, is
proven as follows: suppose that both ˆ
h x and ˆ
h y stabilize ˆ
A 0 ; then their product will
also be a stabilizer:
ˆ
h x ˆ
h y ˆ
A 0 ( ˆ
h x ˆ
h y )
−1 = ˆ
h x ˆ
h y ˆ
A 0 ˆ
h
−1
y
ˆ
h
−1
x = ˆ
h x ˆ
A 0 ˆ
h
−1
x = ˆ
A 0
(3.27)
where we use the result of Eq. (3.10) that the inverse of a product is equal to the
product of the inverses in the reverse order. Next, we expand G in cosets of this
newly found subgroup H . All elements of a coset ˆ
R i H , with ˆ
R i /
∈ H , will transform
ˆ
A 0 into the same new element ˆ
A i :
ˆ
R i ˆ
h x ˆ
A 0 ( ˆ
R i ˆ
h x )
−1 = ˆ
R i ˆ
h x ˆ
A 0 ˆ
h
−1
x
ˆ
R
−1
i = ˆ
R i ˆ
A 0 ˆ
R
−1
i = ˆ
A i
(3.28)
The result must be different from ˆ
A 0 because, otherwise, ˆ
R i w o u l db ea ne l e m e n t
of H . In order to prove the theorem, the following remaining questions have to be
decided. Do different cosets give rise to different similarity transforms? Does one
obtain all elements of a class by finding all transforms of a given starting element?
The answers to both questions are affirmative. For the first question, if ˆ
R i and ˆ
R j
represent different cosets, one should conclude that ˆ
A i = ˆ
A j . Suppose that the opposite is true:
ˆ
R i ˆ
A 0 ˆ
R
−1
i = ˆ
R j ˆ
A 0 ˆ
R
−1
j
ˆ
R
−1
i
ˆ
R j ˆ
A 0 ˆ
R
−1
j
ˆ
R i = ˆ
A 0
ˆ
R
−1
i
ˆ
R j ˆ
A 0
ˆ
R
−1
i
ˆ
R j
−1 = ˆ
A 0
(3.29)
This implies that ˆ
R
−1
i
ˆ
R j stabilizes ˆ
A 0 and thus must belong to H , where it corresponds to, say, ˆ
h z . But one then again has
ˆ
R i ˆ
h z = ˆ
R i ˆ
R
−1
i
ˆ
R j = ˆ
R j
(3.30)
and thus ˆ
R j is a representative of the same coset as ˆ
R i , which contradicts the starting
assumption. Hence, there will be at least as many equivalent elements in the class as
there are cosets of the stabilizing subgroup. Have we then generated the entire class?
Yes, because by going through all the cosets, we run through the entire group. In this
way, we have found all elements that are conjugate to a given one, but, because of
transitivity, this also means that there cannot be other conjugate elements. The oneto-one mapping between conjugate elements and cosets implies that the number in
a class is equal to the number of cosets of the stabilizing subgroup and hence—by
Lagrange’s theorem—must be a divisor of the group order.