374 10 Group Theory
10.2.3 Subgroups
A subset H of the elements of G that forms a group with respect to the same composition
law is described as a subgroup of G. This relationship is denoted using the notation of sets,
H G. Every group G contains two trivial subgroups, one consisting of just the identity
element I and the other consisting of G itself; these are described as improper.One
of the main problems of group theory is the construction and classification of all proper
subgroups of a specified group. The relevant criteria are that H be closed with respect to
multiplication and that inverses are present for each element; the associative property is
satisfied automatically and every subgroup must contain the identity element.
If K is a subgroup of H and H is a subgroup of G, then K is a subgroup of G. Because
every group must contain its subgroups, one can construct the sequence G c H c K c I.
We now derive Lagrange’s divisor theorem, which states that the order of a subgroup
must be a divisor of the order of its parent. This theorem is obviously true for improper
subgroups containing either one or all elements. Suppose that H is a proper subgroup of
G and label its elements H
i
,i 1,h where h is the order of H and H
1
I is the identity
element. Consider an element a G that does not belong to H and construct the left coset
aH aH
i
,i 1,h containing the products of a with each H
i
. This set is not a group
because it does not contain the identity element. Each element of aH is distinct and none
are included in H. (Note that aH
i
H
j
a H
j
H
1
i
H, contrary to the assumption
that a/ H.) If the union of H and aH, both of order h, is not G itself, we select another
element b and construct a new left coset bH of order h that has no overlap with either H or
aH. (If bH
i
aH
j
then b aH
j
H
1
i
aH because H
j
H
1
i
H.) Thus, we can divide G
into a subgroup H and its cosets according to
G H aH bH (10.14)
where each subset contains h elements. If g is the order of G and h is the order of H, then
g mh where m is the index of the subgroup and equal to the number of independent
cosets plus one. Therefore, h is a divisor of g, as stipulated by Lagrange’s theorem.
Once again consider S
3
, where 3 and 2 are divisors of its order. There is one subgroup
of order 3, namely I,a,b, that is isomorphic to the group of invariant proper rotations of
an equilateral triangle, and there are three subgroups of order 2, namely I,c, I,d, I,e
based upon reflections across a bisector.
If a G is an element of order k, we can construct a cyclic group I,a,a
2
,...,a
k1
where a
k
I. This cyclic group, called the period of a, is the smallest subgroup of G that
contains a and its order must be a divisor of the order of G. Therefore, all groups with
prime order must be cyclic and can be generated from one of its elements other than the
identity.
Although our derivation of the divisor theorem employed left cosets aH, we could just
as easily have chosen right cosets Ha. Left and right cosets need not be the same, but the
argument still works. In the special case that aH Ha H aHa
1
for any a G,we
describe H as an invariant subgroup of G. Invariant subgroups clearly must contain one
or more complete equivalence classes. More generally, if H is a subgroup of G, and a is
any element of G, then aHa
1
is also a subgroup of G described as conjugate (or similar)