Why Read This Book? - Index of

8.3 Quotient Groups 261
[0] ={...,−2n, −n, 0,n,2n,...}
[1] ={...,−2n + 1, −n + 1, 1,n+ 1, 2n + 1,...}
[2] ={...,−2n + 2, −n + 2, 2,n+ 2, 2n + 2,...}
.
[k] ={...,−2n + k, −n + k, k, n + k, 2n + k,...}
.
[n − 1] ={...,−n − 1, −1,n− 1, 2n − 1, 3n − 1,...}
(8.33)
Notice the very important fact that [0] =(n); that is, the equivalence class
of the identity in (Z, +, 0, −) is the subgroup from which the equivalence
is defined. Lump these n equivalence classes into a family and call it Z/(n),
the integers modulo n.
Z/(n) ={[0], [1], [2],...,[n − 1]} (8.34)
Recall that every equivalence class has infinitely many names, depending
on the representative element by which we choose to address it. For
example, in Z/(6), [5] =[−1] =[41].
Step 2: Define a binary operation on the created set. On the elements of Z/(n),
which are themselves sets, we define a form of addition ⊕n in the following
way:
[a]⊕n [b] =[a +Z b] (8.35)
where +Z denotes the sum of the representative integers a and b. For
example, [4]⊕6 [5] =[4 + 5] =[9] =[3], where we choose to refer to the
sum as [3] since [3] is the same set as [9] and 3 is a representative element
between 0 and 5.
Notice that this new binary operation ⊕n is a way of combining two sets
to produce another set. It is not union or intersection, which up to now
are the only binary operations on sets we have seen. Instead, ⊕n combines
[a] and [b] by using representative elements from each to produce
a representative element of the set we are defining to be [a]⊕n [b].
Step 3: Show that the created set with its binary operation gives rise to a group.
We must show that ⊕n is well defined, closed, and associative. We must
also find an identity element and find inverses for all elements. Clearly, ⊕n
is closed. For if [a], [b] ∈Z/(n), then a + b is an integer. Since the equivalence
classes in Z/(n) partition the integers, a + b is in some equivalence
class. Thus [a + b] ∈Z/(n).