Section 9.1 Algebraic Structures
699
Suppose that A = E, the set of even integers. E is closed under addition, E
contains 0 (the identity element), and the inverse of every even integer (its negative) is an even integer. 3E, + 4 is thus a group. But suppose that A = O, the set of
odd integers. 3O, + 4 fails to be a group for several reasons. For one thing, it is not
closed—adding two odd integers produces an even integer. (Closure depends on
the set as well as the operation, so it is not an inherited property). For another, a
subgroup must have an identity with respect to addition; 0 is the only integer that
will serve, and 0 is not an odd integer.
Definition suBGrouP
Let 3G, # 4 be a group and A # G. Then 3A, # 4 is a subgroup of 3G, # 4 if 3A, # 4 is
itself a group.
For 3A, # 4 to be a group, it must have an identity element, which we’ll denote
by i A . Of course G also has an identity element, which we’ll denote by i G . It turns
out that i A = i G , but this equation does not follow from the uniqueness of a group
identity because the element i A , as far as we know, may not be an identity for all of
G, and we cannot yet say that i G is an element of A. However, i A = i A # i A because
i A is the identity for 3A, # 4, and i A = i A # i G because i G is the identity for 3G, # 4. Because of the left cancellation law holding in the group 3G, # 4, it follows that i A = i G .
To test whether 3A, # 4 is a subgroup of 3G, # 4, we can assume the inherited
properties of a well-defined operation and associativity, and we check for the three
remaining properties required.
tHeoRem on suBGrouPs
For 3G, # 4 a group with identity i and A # G, 3A, # 4 is a subgroup of 3G, # 4 if it
meets the following three tests:
1. A is closed under # .
2. i [ A.
3. Every x [ A has an inverse element in A.
PrACtiCe 22 The definition of a group requires that the set be nonempty. In the theorem on subgroups, why isn’t there a specific test that A ∙ [?
example 13
a. 3Z, + 4 is a subgroup of the group 3R, + 4. Z is closed under addition, 0 [ Z,
and the negative of every integer is an integer.
b. 3 51,46, # 5 4 is a subgroup of the group 3 51, 2, 3, 46, # 5 4. Closure holds:
# 5 1 4
1
1
4
4
4
1
The identity 1 [ 51, 46, and 1
−1
= 1, 4
−1
= 4).
■
699
Suppose that A = E, the set of even integers. E is closed under addition, E
contains 0 (the identity element), and the inverse of every even integer (its negative) is an even integer. 3E, + 4 is thus a group. But suppose that A = O, the set of
odd integers. 3O, + 4 fails to be a group for several reasons. For one thing, it is not
closed—adding two odd integers produces an even integer. (Closure depends on
the set as well as the operation, so it is not an inherited property). For another, a
subgroup must have an identity with respect to addition; 0 is the only integer that
will serve, and 0 is not an odd integer.
Definition suBGrouP
Let 3G, # 4 be a group and A # G. Then 3A, # 4 is a subgroup of 3G, # 4 if 3A, # 4 is
itself a group.
For 3A, # 4 to be a group, it must have an identity element, which we’ll denote
by i A . Of course G also has an identity element, which we’ll denote by i G . It turns
out that i A = i G , but this equation does not follow from the uniqueness of a group
identity because the element i A , as far as we know, may not be an identity for all of
G, and we cannot yet say that i G is an element of A. However, i A = i A # i A because
i A is the identity for 3A, # 4, and i A = i A # i G because i G is the identity for 3G, # 4. Because of the left cancellation law holding in the group 3G, # 4, it follows that i A = i G .
To test whether 3A, # 4 is a subgroup of 3G, # 4, we can assume the inherited
properties of a well-defined operation and associativity, and we check for the three
remaining properties required.
tHeoRem on suBGrouPs
For 3G, # 4 a group with identity i and A # G, 3A, # 4 is a subgroup of 3G, # 4 if it
meets the following three tests:
1. A is closed under # .
2. i [ A.
3. Every x [ A has an inverse element in A.
PrACtiCe 22 The definition of a group requires that the set be nonempty. In the theorem on subgroups, why isn’t there a specific test that A ∙ [?
example 13
a. 3Z, + 4 is a subgroup of the group 3R, + 4. Z is closed under addition, 0 [ Z,
and the negative of every integer is an integer.
b. 3 51,46, # 5 4 is a subgroup of the group 3 51, 2, 3, 46, # 5 4. Closure holds:
# 5 1 4
1
1
4
4
4
1
The identity 1 [ 51, 46, and 1
−1
= 1, 4
−1
= 4).
■