Algebra I: Section 6. The structure of groups. 6.1 Direct products of ...

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Figure 6.1 The product group Zm × Zn represented as an m × n array. The
element 1 = (1, 1) and its iterates are shaded. In this portrayal the array is
7 × 12; since gcd(7, 12) = 1 the iterates will cycle once through each point in the
array before returning to position 1.
so that [k]mn = [ℓ]mn, as required to show that ψ is one-to-one. The map ψ is the desired
isomorphism between groups.
To connect all this with the remainder problem, if m, n are relatively prime the element
1 = ([1]m, [1]n) must be a cyclic generator for Zm × Zn because it is the ψ-image of the
generator [1]mn in the cyclic group Zmn. Given a1, a2, consider the element a = ([a1]m, [a2]n)
in the product group. Then some multiple of 1 is equal to a, say a = k · 1. Then
([a1]m, [a2]n) = a = k · 1 = (k · [1]m, k · [1]n) = ([k]m, [k]n)
so that k ≡ a1 (mod m) and k ≡ a2 (mod n), which is the same as saying that x0 = k is
a solution of the congruence (5). If k ′ is another solution then k ′ · 1 = a = k · 1 so we get
(k ′ − k) ·1 = a −a = 0 = ([0]m, [0]n). But 1 has order mn, being the generator of Zmn, so this
happens ⇔ (k ′ − k) is a multiple of mn. The full set of solutions is therefore x0 + Z · mn as
claimed. �
The converse of this theorem is not true. For example, gcd(2, 4) > 1 and the direct product
Z4 × Z2 is not isomorphic to Z8: the maximal order of any element x in the product group is
4 because 4 · ([k]4, [ℓ]2) = (4[k]4, 4[ℓ]2) = ([0]4, [0]2). But Z8 has a cyclic generator of order 8.
6.1.27 Exercise. Is Z45 ∼ = Z9 × Z5? Is Z45 ∼ = Z15 × Z3? Prove or explain why not.
Hint: Compare orders of elements. �
6.1.28 Exercise. In Z7 × Z12 there is some exponent k such that
k · 1 = k · ([1]7, [1]12) is equal to ([3]7, [−4]12)
Find a k that does this. Then find a “normalized ” solution lying in the range 0 ≤ k < 7·12 = 84.
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The underlying idea of this proof is illustrated in Figure 6.1 where we represent Zm×Zn as an
m×n array of squares. Counting from the bottom left square 0 = (0, 0), the element 1 = (1, 1) is
the dark shaded square and its “powers” 1,1+1 = (2, 2), . . . move diagonally upward until they
hit an edge, at which point they re-enter the array from the opposite edge. If gcd(m, n) = 1 all
squares get hit exactly once until we finally arrive back at 0 = mn ·1; the pattern then repeats.
This observation tells us how to compute the inverse ψ −1 : Zm ×Zn → Zmn of our isomorphism.
A pair a = ([a1]m, [a2]n) maps back to [k]mn in Zmn ⇔ k is the first integer k = 0, 1, 2, . . . such
that k ·1 = a. This k is of course the desired solution of the congruence system (5), so knowing
how to compute ψ −1 is equivalent to being able to solve the remainder problem.
Systematic solution of remainder problems. Start by working backward: if x is a solution to
(5) then there exist k, ℓ ∈ Z such that x = a1 + km = a2 + ℓn, and hence a2 − a1 = km − ℓn.
Conversely, given k, ℓ satisfying the latter identity we could recover x as a1 + km or a2 + ℓn,
which automatically have the same value. So, we procede as follows
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