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

6.1.18 Exercise (Groups of order p 2 ). Assume G is a finite group of order |G| = p 2 for
some prime p > 1. Prove that G ∼ = Z p 2 or G ∼ = Zp × Zp.
Hint: Use the Cauchy theorems and adapt the ideas developed in 6.1.17. �
6.1.19 Exercise. Prove that |A1 × . . . × Ar| = |A1| · . . . · |Ar| for any direct product of groups.
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6.1.20 Exercise. Prove that the order o(x) of an element x = (a, b) in the direct product
group A × B is the least common multiple lcm(o(a), o(b)). Is a similar result true for direct
products of several groups?
Hint: Observe that x m = e ⇔ a m = e and b m = e. In any group, g m = e ⇔ m is a multiple of
o(g). �
6.1.21 Exercise. Is Z15 × Z4 a cyclic group – i.e. does it have an element of order 60? Does
Z15 × Z5 have elements of order (i) 75? (ii) 25? (iii) 15? (iv) 3? (v) any other order? �
6.1.22 Exercise. Identify all elements x = (a, b) of order o(x) = 5 in Z15 × Z5. Same for
Z15 × Z4. �
Ultimately we will have a lot to say about the structure of a finite group G in terms of the
prime divisors of its order n = |G|. There is one important case in which the outcome has a
striking simplicity.
6.1.23 Theorem. Let G be a nontrivial finite group of prime order |G| = p > 1. Then G is
isomorphic to the cyclic group (Zp, +). Furthermore, every b �= e is a cyclic generator.
Proof: Let b �= e and consider the cyclic subgroup H = {e, b, b 2 , . . . , b k−1 }, with b k = e. By
Lagrange, |H| = k ≥ 2 must divide the order |G| = p of the whole group. Hence k = p and
H = G. �
Direct products and the Chinese Remainder Theorem. The Chinese Remainder Theorem
(CRT) has its roots in number theory but has many uses. One application completely
resolves the issues regarding direct products Zm × Zn mentioned in 6.1.14 - 16. The original
remainder theorem arose in antiquity when attempts were made to solve systems of congruences
involving several different moduli ni
x ≡ a1
(mod n1)
. x ∈ Z
x ≡ ar
(mod nr)
The notion of congruence is a modern one; the ancient chinese would have viewed this problem
as the search for an integer x with specified remainders ai after division by ni, i = 1, 2, . . .r.
Such systems do not always have solutions, but solutions do exist if the moduli n1, . . . , nr
are pairwise relatively prime, so that gcd(ni, nj) = 1 if i �= j, and then the solution x is unique
up to added multiples of the least common multiple m = lcm(n1, . . .,nr) of the moduli.
6.1.24 Exercise. Here are two systems of congruences
(a)
�
x ≡ 5
x ≡ 1
(mod 3)
(mod 12)
(b)
� x ≡ 5 (mod 3)
x ≡ 1 (mod 5)
Taking a “bare hands” approach, verify that the system (a) has no solutions and that the
solutions of (b) are of the form x0 + k · 15 where x0 = 11 and k ∈ Z.
Hint: If m, n > 1 recall our discussion in Chapter 2 where we showed that the greatest common
divisor c = gcd(m, n) is the smallest positive element in the lattice Λ = Zm+Zn. In particular:
(i) there exist integers r, s (which are not hard to find by trial and error) such that gcd(m, n) =
rm+sn, and (ii) the elements in Λ are precisely the integer multiples of gcd(m, n). The moduli
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