242 A Introduction to Group Theory
• It is seen that, except for the identity e, the permutations leave no symbol un-
changed since π
b
takes a
i
into ba
i
, which is equal to a
i
only if b is the identity e.
Permutations in S
n
with these two properties are called regular permutations.
2Cosets
Lagrange’s theorem The order of a subgroup of a finite group is a divisor of the
order of the group.
Suppose two sets of h distinct elements each, H and aH contained in G.IfG has
not been exhausted, choosing some element b of G which is not contained in either
H or aH. Thus, the set bH will again generate h new elements of the group G.
Continuing this process, we can exhibit the group G as the sum of a finite number
of distinct sets of h elements each
G = H +aH +bH +···+mH. (A.7)
Thus the order g of the group G is a multiple of the order h of its subgroup H , i.e.,
g =mh, (A.8)
where m is the index of the subgroup H under the group G. This means that h is
a divisor of g so that the orders of all elements of a finite group must be divisors
of the order of the group. The sets of elements of the form aH in (A.7) are called
left cosets of H in G. Of course, we are able to multiply H on the right to yield the
corresponding right cosets as follows
G =H +Ha
1
+Hb
1
+···+Hm
1
. (A.9)
It is seen from Eq. (A.8) that a group, whose order is a prime number, has no proper
subgroups and is necessarily cyclic. Such a group can be generated from any of its
elements other than the identity element.
Lagrange’s theorem can be used to find the possible structures of groups for a
given order. Here we present an explicit example, say order 6, to show its advan-
tage. Since the order of the group is 6, then the order of each of its elements is
a divisor of 6, i.e., 1, 2, 3 or 6. If the group contains an element a of order 6,
then the group is the cyclic group {a,a
2
,a
3
,a
4
,a
5
,a
6
=e}. To find other possible
structures, let us assume that the group contains no element of order 6, but has an
element a of order 3. Thus this group contains the subgroup H ∈{a,a
2
,a
3
= e}.
If the group also contains another element b, then it contains six different elements
{e, a, a
2
,b,ba,ba
2
}. The element b has order 2 or 3. After analyzing its order care-
fully, it is found that the order of element b cannot be 3 and it must be 2. This is
because b
3
= e if the order of b is 3, then element b
2
must be one of the six ele-
ments listed above. Obviously, we cannot have b
2
=e. Assuming that b
2
=b, ba or
ba
2
implies that b =e, a or a
2
, which contradicts the assumption that b is different
from these elements. Moreover, if we assume that b
2
= a or a
2
implies ba = e or