Section 9.1 Algebraic Structures
701
The order of the group 3S n , + 4 (the number of elements) is n! What is the order
of the subgroup 3A, + 4? We might expect half the permutations in S n to be even and
half to be odd. Indeed, this is the case. If we let O n denote the set of odd permutations in S n (which is not closed under function composition), then the mapping
f : A n S O n defined by f (a) = a + (1, 2) is a bijection.
PrACtiCe 25 Prove that f : A n S O n , given by f (a) = a + (1, 2) is one-to-one and onto.
■
Because there is a bijection from A n onto O n , each set has the same number of
elements. But A n d O n = [ and A n c O n = S n , so 0 A n 0 = 0 S n 0 ∙2 = n!∙2.
tHeoRem on ALtErnAtinG GrouPs
For n [ N, n > 1, the set A n of even permutations determines a subgroup, called
the alternating group, of 3S n , + 4 of order n!∙2.
We have now seen several examples of subgroups of finite groups. In Example
13b and Practice 23, there were three such examples, and the orders of the groups
and subgroups were
Group of order 4, subgroup of order 2
Group of order 8, subgroup of order 4
Group of order 6, subgroup of order 3
The theorem on alternating groups says that a particular group of order n! has a
subgroup of order n!∙2.
Based on these examples, one might conclude that subgroups are always half
the size of the parent group. This conclusion is not always true, but there is a
relationship between the size of a group and the size of a subgroup. This relationship is stated in Lagrange’s theorem, proved by the great French mathematician
Joseph-Louis Lagrange in 1771 (we will omit the proof here).
tHeoRem LAGrAnGE’s thEorEM
The order of a subgroup of a finite group divides the order of the group.
Lagrange’s theorem helps us narrow down the possibilities for subgroups of
a finite group. If 0 G0 = 12, for example, we would not look for any subgroups of
order 7 because 7 does not divide 12. Also, the fact that 6 divides 12 does not
imply the existence of a subgroup of G of order 6. In fact, A 4 is a group of order
4!∙2 = 12, but it can be shown that A 4 has no subgroups of order 6. Therefore the
converse to Lagrange’s theorem does not always hold. In certain cases the converse can be shown to be true—for example, in finite commutative groups (note
that A 4 is not commutative).
Finally, we consider subgroups of the group 3Z, + 4. For n any fixed element
of N, the set nZ is defined as the set of all integral multiples of n; nZ = 5nz 0 z [ Z6.
Thus, for example, 3Z = 50, ± 3, ± 6, ± 9, … 6.
701
The order of the group 3S n , + 4 (the number of elements) is n! What is the order
of the subgroup 3A, + 4? We might expect half the permutations in S n to be even and
half to be odd. Indeed, this is the case. If we let O n denote the set of odd permutations in S n (which is not closed under function composition), then the mapping
f : A n S O n defined by f (a) = a + (1, 2) is a bijection.
PrACtiCe 25 Prove that f : A n S O n , given by f (a) = a + (1, 2) is one-to-one and onto.
■
Because there is a bijection from A n onto O n , each set has the same number of
elements. But A n d O n = [ and A n c O n = S n , so 0 A n 0 = 0 S n 0 ∙2 = n!∙2.
tHeoRem on ALtErnAtinG GrouPs
For n [ N, n > 1, the set A n of even permutations determines a subgroup, called
the alternating group, of 3S n , + 4 of order n!∙2.
We have now seen several examples of subgroups of finite groups. In Example
13b and Practice 23, there were three such examples, and the orders of the groups
and subgroups were
Group of order 4, subgroup of order 2
Group of order 8, subgroup of order 4
Group of order 6, subgroup of order 3
The theorem on alternating groups says that a particular group of order n! has a
subgroup of order n!∙2.
Based on these examples, one might conclude that subgroups are always half
the size of the parent group. This conclusion is not always true, but there is a
relationship between the size of a group and the size of a subgroup. This relationship is stated in Lagrange’s theorem, proved by the great French mathematician
Joseph-Louis Lagrange in 1771 (we will omit the proof here).
tHeoRem LAGrAnGE’s thEorEM
The order of a subgroup of a finite group divides the order of the group.
Lagrange’s theorem helps us narrow down the possibilities for subgroups of
a finite group. If 0 G0 = 12, for example, we would not look for any subgroups of
order 7 because 7 does not divide 12. Also, the fact that 6 divides 12 does not
imply the existence of a subgroup of G of order 6. In fact, A 4 is a group of order
4!∙2 = 12, but it can be shown that A 4 has no subgroups of order 6. Therefore the
converse to Lagrange’s theorem does not always hold. In certain cases the converse can be shown to be true—for example, in finite commutative groups (note
that A 4 is not commutative).
Finally, we consider subgroups of the group 3Z, + 4. For n any fixed element
of N, the set nZ is defined as the set of all integral multiples of n; nZ = 5nz 0 z [ Z6.
Thus, for example, 3Z = 50, ± 3, ± 6, ± 9, … 6.