Why Read This Book? - Index of

256 Chapter 8 Groups
a cyclic subgroup. The easiest way to see what (a) looks like is to build it from the
bottom up, so to speak, and then demonstrate that what you have created satisfies
U1–U3. Using the definition of a n in Eqs. 8.12–8.14, consider the set
S ={a n : n ∈ Z} (8.22)
The claim is that S = (a), which we verify here by showing that S has properties
U1–U3, thereby earning the right to be called (a). The details will help you in
Exercise 8.2.24.
Let n = 1 to see that a ∈ S, so that S has property U1. To show S has
property U2, we must show it has properties H1–H3.
(H1) Pick x, y ∈ S. Then there exist integers m and n such that x = am and
y = an . Now m + n is also an integer, so am ∗ an = am+n ∈ S. Thus S is
closed under ∗.
(H2) Letting n = 0, we have that e = a0 ∈ S.
(H3) Pick an ∈ S. Since Theorem 8.1.24 applies for all integers n, we have that
(an ) −1 = a−n ∈ S. Thus S is closed under inverses.
To show that S has property U3, suppose B is a subgroup and a ∈ B. We show
S ⊆ B by showing a n ∈ B for all n. Since a ∈ S and B is closed under ∗, it must be
that a n ∈ B for all positive integers n. Certainly, a 0 = e ∈ B, and since B is closed
under inverses, a −n ∈ B for all positive integers n. Thus S ⊆ B, and we have
finished the proof that (a) ={a n : n ∈ Z}.
Example 8.2.14 In the multiplicative group of nonzero real numbers,
(3) ={3 n : n ∈ Z} ={...,1/27, 1/9, 1/3, 1, 3, 9, 1/27,...} �
Example 8.2.15 We construct (3) in the additive group (Z, +, 0, −). Note that
zero is the identity in this context, so that 3 0 = 0. Constructing 3 n for n ≥ 2is
repeated addition, not repeated multiplication. Thus
3 1 = 3, 3 2 = 3 + 3 = 6, 3 3 = 6 + 3 = 9, 3 4 = 9 + 3 = 12, etc. (8.23)
Constructing 3 −n involves negation, not reciprocation, so that
3 −1 =−3
Thus (3) ={3k : k ∈ Z}. �
3 −2 = (3 −1 ) 2 =[(−3) + (−3)] =−6
3 −3 = (3 −1 ) 3 =[(−6) + (−3)] =−9, etc.
(8.24)