(a) Because K is the kernel of the homomorphism G ! G0, f ((ki kj))¼ f(ki) f (kj)¼ E0E0¼ E0. Therefore, kikj2 K. The set K is therefore closed and so K is a subgroup of G. Consider the mapping of gjkigj1, ki2 K, gj2 G,
fðgjkig1j Þ ¼ f ðgjÞ f ðkiÞ f ðgj1Þ ¼ f ðgiÞE0fðg1j Þ ¼ E0, (1) where we have used eq. (1.5.7) and Exercise1.5-1. Therefore,
(1) gjkigj12 K, (2)
which shows K¼ {ki} to be an invariant subgroup of G.
(b) The subgroup C(3) is the kernel of S(3) for the homomorphism of S(3) on to its factor group F because f (C(3))¼ E0.
(c) No two fibers in G can have a common element; otherwise this common element would have two distinct images in G0, which is contrary to the requirements for a mapping.
Therefore, there are as many disjoint fibers in G as there are elements in G0, namely g0. It remains to be shown that all fibers in G have the same order, which is equal to the order k of the kernel K. Firstly, the necessary and sufficient condition for two elements g2, g3that are 2 G to belong to the same fiber of G is that they be related by
g2¼ g3ki, ki2 K: (3)
Sufficiency:
fðg2Þ ¼ f ðg3kiÞ ¼ f ðg3Þ f ðkiÞ ¼ f ðg3ÞE0¼ f ðg3Þ: (4) Necessity: Suppose that g2¼ g3gj; then f (g2)¼ f (g3), f (gj). But if g2, g3belong to the same fiber then f(gj) must be E0and so gjcan only be2 K. Secondly, if gnis a particular element of a fiber Fn, then the other elements of Fncan all be written in the form gnki, where ki2 K,
Fn¼ fgn kig, ki2 K: (5)
All the distinct elements of Fnare enumerated by eq. (5) as i¼ 1, 2, . . . , k, the order of K.
Therefore, the number of elements in each one of the g0fibers in G is k, whence the order of G is
g¼ k g0, (6)
which establishes the required result.
1.8 More about subgroups and classes
If G and H are two groups for which a multiplication rule exists, that is to say the result gihj
is defined, then the conjugate of H by an element gi2 G is giH gi1¼P
j
gihjg1i : (1)
When the result is H itself, H is invariant under the element gi,
gi H g1i ¼ H: (2)
(2) gi H¼ H gi, (3)
which is an equivalent condition for the invariance of H under gi. The set of elements {gi}2 G that leave H invariant form a subgroup of G called the normalizer of H in G, written N(H|G). That N(H|G) does indeed form a subgroup of G follows from the fact that if gi, gj2 N(H|G)
gj H g1j ¼ H, (4)
(2), (4) gi gj HðgigjÞ1 ¼ gi H g1i ¼ H, (5)
(5) gi gj2 fgi, gj, . . .g ¼ NðHjGÞ, (6)
implying closure of {gi, gj, . . . }, a sufficient condition for {gi, gj, . . . } to be a subgroup of G. If the normalizer N(H|G) is G itself, so that H is invariant under all gi2 G, H is said to be normal or invariant under G. If H is a subgroup of G (not so far assumed) then H is an invariant subgroup of G if eqs. (2) and (3) hold.
If G, H are two groups for which a multiplication rule exists then the set of all the elements of G that commute with a particular element hjof H form a subgroup of G called the centralizer of hjin G, denoted by
ZðhjjGÞ G: (7)
H may be the same group as G, in which case hjwill be one element of G, say gj2 G.
Similarly, the centralizer of H in G,
ZðHjGÞ G, (8)
is the set of all the elements of G that each commute with each element of H; H in eq. (8) may be a subgroup of G. If H is G itself then
ZðGjGÞ ZðGÞ (9)
is the center of G, namely the set of all the elements of G that commute with every element of G. In general, this set is a subgroup of G, but if Z(G)¼ G, then G is an Abelian group.
Exercise 1.8-1 Prove that the centralizer Z(hj|G) is a subgroup of G.
Exercise 1.8-2 (a) Find the center Z(C(3)) of C(3). (b) What is the centralizer Z(C(3)|S(3)) of C(3) in S(3)? (c) What is the centralizer Z(P1|S(3)) of P1in S(3)?
A class was defined in Section 1.2 as a complete set of conjugate elements. The sum of the members gj(ci), j¼ 1, 2, . . . , ciof the class cithat contains the group element giis
1.8 More about subgroups and classes 19
ðciÞ ¼P
j
gjðciÞ, (10)
ci¼ fgkgig1k g, 8 gk2 G, (11) with repetitions deleted. The sum of all the elements in a class, (ci), is the Dirac character of the class ci, and
(10), (11) ðciÞ ¼P
k
gkgigk1, (12)
with repetitions deleted. It is rather a waste of effort to evaluate the transforms on the right side (RS) of eq. (12) for all gk2 G, since many redundancies will be found that will have to be eliminated under the ‘‘no repetitions’’ rule. For instance, see Example1.2-1, where six transforms of P1yield a class that contains just two members, P1and P2, each of which occurred three times. However, it is possible to generate the class cithat contains gi without redundancies, from the coset expansion of G that uses the centralizer of gi
as the subgroup in the expansion. Abbreviating Z(gi|G) to Zi, the coset expansion of G on Ziis
G¼Pt
r¼1
gr Zi, g1¼ E, t ¼ g=z, (13)
where z is the order of Zi. From the definition of the coset expansion in eq. (13), the elements of {gr} with r¼ 2, . . . , t, and Z are disjoint. (E is of course 2 Zi.) We shall now prove that
ðciÞ ¼P
r
grgig1r , (14)
where {gr} is the set defined by eq. (13), namely the t coset representatives.
Proof The coset expansion eq. (13) shows that G¼ {gk} is the DP set of {zp} and {gr}, which means that G may be generated by multiplying each of the z members of {zp} in turn by each of the t members of {gr}. Therefore, gkin eq. (12) may be written as
gk ¼ zpgr, gk2 G, zp2 Zi, (15) with {gr} defined by eq. (13). In eq. (15), p, which enumerates the z elements of Zi, runs from 1 to z; r, which enumerates the coset representatives (including g1¼ E), runs from 1 to t;
and k enumerates all the g elements of the group G as k runs from 1 to g.
(12), (15) P definition of the centralizer, all commute with gi. The third equality follows because the double sum consists of the same t terms repeated z times as p runs from 1 to z. It follows
from the uniqueness of the binary composition of group elements that the sum over r in eq. (16) contains no repetitions. Therefore the sum over r on the RS of (16) is (ci), which establishes eq. (14). Since eq. (14) gives the elements of ciwithout repetitions, the order ci
of this class is
ci t ¼ g=z: (17)
Equation (17) shows that the order of a class ci is a divisor of the order of the group (Lagrange’s theorem). It also yields the value of cionce we determine z from Zi Z(gi|G).
The t elements grneeded to find the Dirac character (ci) of the class ci, and thus the members of ci, are the coset representatives of the centralizer Zi Z(gi|G).
Exercise 1.8-3 Find the class of P1in S(3) by using the coset expansion for the centralizer Z(P1|S(3)) and eq. (14).