Why Read This Book? - Index of

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8.1 Introduction to Groups 249
(8.5)
To verify that the binary operation in Exercise 8.1.19 is associative would be
a formidable task if all you have is the Cayley table. Suffice it to say that ◦ is
associative, so that S is a non-abelian group. This group is actually motivated by a
group whose elements are functions. You will see this group in Section 8.4.
If someone gives you a set with a binary operation and asks you to show that
it is a group, you must verify properties G1–G5. To give you an idea of how that
might look, we will now walk through most of the details of a specific example.
Define the complex numbers by
C ={a + bi : a, b ∈ R} (8.6)
At this point, i is a mere symbol with no meaning. We should think of C merely as
a set of ordered real number pairs, the first of which stands alone and the second
of which is tagged with an adjacent i. Furthermore, the + in a + bi is not meant to
denote real addition, as if the expression a + bi could be simplified. To avoid this
possible confusion, some authors define elements of C as real number ordered
pairs (a, b).
Next we define equality in C, which for clarity we temporarily denote ≡. We
define a1 + b1i ≡ a2 + b2i, provided the real number equations a1 = a2 and b1 =
b2 are satisfied. To show that ≡ is an equivalence relation is pretty trivial. For
example, to show ≡ has property E2, suppose a1 + b1i ≡ a2 + b2i. Then a1 = a2
and b1 = b2. Since real number equality has property E2, we have a2 = a1 and
b2 = b1. Thus a2 + b2i ≡ a1 + b1i.
Define the binary operation ⊕ in the following way.
(a + bi) ⊕ (c + di) = (a + c) + (b + d)i (8.7)
We claim that C is an abelian group under ⊕. Here are the steps of the proof in
meticulous detail. Working through them will give you a good sense of direction in
the exercise that follows, where you will show that the nonzero complex numbers
C × with a form of multiplication is a group.
(G1) Suppose a1 + b1i ≡ a2 + b2i and c1 + d1i ≡ c2 + d2i. Then a1 = a2,
b1 = b2, c1 = c2, and d1 = d2. Since real number addition is well defined,