The group (Z/nZ) February 17, 2016 1 Introduction In these notes we figure out the structure of the unit group (Z/nZ) where n > 1 is an integer. If we factor n = p e 1 1 pe, where the p i s are distinct primes, then since gcd(p e i Chinese Remainder Theorem guarantees that i, pe j j Z/nZ = Z/p e 1 1 Z... Z/pe Z ) = 1 for i j, the as rings. Taing the unit groups of both sides of this isomorphism yields (Z/nZ) = (Z/p e 1 1 Z)... (Z/p e Z). This shows us that in order to understand the structure of (Z/nZ) it suffices to understand the structure of (Z/p e Z) where p is a prime and e 1 is an integer. For a positive integer n, we define ϕ(n) := (Z/nZ), then the above isomorphism shows that ϕ(n) = i=1 ϕ(p e i i ). A positive integer a p e has a multiplicative inverse in Z/p e Z if and only if 1 = gcd(a, p e ) = gcd(a, p). Thus ϕ(p e ) equals the number of positive integers less than p e which are relatively prime to p. Of course gcd(a, p) 1 1
if and only if p a, so the number of elements in Z/p e Z without inverses equals p e p = pe 1. Thus In this document we will prove ϕ(p e ) = p e p e 1 = p e 1 (p 1). Theorem 1.1. Let p be a prime and e 1 an integer. Then 1. (Z/p e Z) is cyclic of size (p 1)p e 1 if p is odd. 2. (Z/2 e Z) is isomorphic to the direct product of a cyclic group of order 2 and a cyclic group of order 2 e 2 when e 2. 2 Preliminaries Before getting to the main theorem, we need to better understand properties of abelian groups. Let G be an abelian group, then we have Proposition 2.1. Suppose G contains an element of order a 1 and an element of order b 1. Then G contains an element of order lcm(a, b). Proof. Let x G have order a and y G have order b. Write a = p e 1 1 pe 2 2 pe b = p f 1 1 pf 2 2 pf where the p i s are prime and e i, f i 0. Then we have the formula lcm(a, b) = p g 1 1 pg 2 2 pg where g i = max(e i, f i ). Reorder the p i s so that e i f i for 1 i j and e i < f i for j + 1 i and define m = p e 1 1 pe 2 2 pe j j n = p f j+1 j+1 pf. Then gcd(m, n) = 1 and mn = lcm(a, b). By definition, m a and n b. Thus r = x a/m G has order m and s = y b/n has order n. I claim that rs has order mn. To prove this, first note that (rs) mn = r m s n = x a y b = 1 2
showing that rs has finite order and that its order is at most mn. For the converse, let t denote the order of rs. Then 1 = (rs) tm = (r m ) t (s tm ) = s tm. Since the order of n is n, we see that n tm but gcd(m, n) = 1 implies n t. A similar argument with tn in place of tm shows that m t, and since gcd(m, n) = 1 we must have mn t hence mn t. Thus t = mn and hence G contains an element of order mn = lcm(a, b). As a corollary of this proposition, we obtain the following result for finite abelian groups. Corollary 2.2. Let G be a finite abelian group. Suppose that M is the maximal order of any element in G. Then a M = 1 for every a G. Proof. Assume for the sae of contradiction that there exists x G with x M 1. If t denotes the order of x, then certainly t does not divide M. By definition of M we now that G contains an element of order M and hence by Proposition 2.1 we see that G contains an element of order lcm(t, M) > M. This contradicts the maximality assumption on M, and therefore a M = 1 for every a G. With Corollary 2.2 on hand we can now prove that the unit group of any finite field is cyclic. Theorem 2.3. Let F be a finite field. Then F is a cyclic group under multiplication. Proof. We now that F is a finite abelian group under multiplication. Let M be the maximal order of an element in F. Then by Corollary 2.2 we now that a M = 1 for every a F. This implies that the polynomial x M 1 F [x] has F many roots. But x M 1 can not have more than M roots, so in fact F M. By Lagrange we now that M F and hence we get equality. Since the maximal order of an element in F equals F, we see that in fact F must be a cyclic group. An immediate corollary that we will use is 3
Corollary 2.4. The group (Z/pZ) is cyclic for any prime p. Proof. F p = Z/pZ is a field. This will be our base case in trying to find the structure of (Z/p e Z) in the next sections. 3 The case p is odd In this section we will prove that (Z/p e Z) is cyclic for all e 1. We already now this result in the case e = 1 since Z/pZ is a field. Notice that Z/p e Z is NOT a field if e > 1, so we must use other techniques. The basic idea here is the following. We ve already computed that ϕ(p e ) = p e 1 (p 1). Thus to show that (Z/p e Z) is cyclic, we must produce an element of order p e 1 (p 1). Since gcd(p e 1, p 1) = 1, Proposition 2.1 tells us that this will be accomplished once we produce elements of order p e 1 and p 1. Proposition 3.1. The group (Z/p e Z) contains an element of order p 1. Proof. In Z, the ideal (p) contains the ideal (p e ). Thus by the third isomorphism theorem, Z/pZ = (Z/p e Z)/(pZ/p e Z). Thus there is a surjective quotient map π e : Z/p e Z Z/pZ. Let g Z/pZ be a generator for the cyclic group (Z/pZ), and let h Z/p e Z satisfy π e (h) = g. If t is the order of h, then 1 = π e (h t ) = π e (h) t = g t. Since the order of g is p 1, we see that p 1 t. Therefore, the order of h t/(p 1) in Z/p e Z is p 1. Proposition 3.2. For p an odd prime, the order of 1 + p in Z/p e Z is p e 1. Proof. We will prove by induction on the stronger statement (1 + p) p = 1 + p +1 t where t Z is relatively prime to p. This is clearly true for = 0 with t = 1. Assume that we now (1 + p) p = 1 + p +1 t 4
for some 0 where gcd(p, t) = 1. By binomial expansion we have (1 + p) p+1 = (1 + p +1 t) p = p i=0 ( ) p (p +1 t) i. i We now that 3+3 +3 for any 0. Thus modulo p +3 this expression equals p i=0 ( ) p (p +1 t) i = 1 + p p +1 p(p 1) t + p 2+2 t 2 (mod p +3 ). i 2 = 1 + p +2 t + p 2+3 t 2 p 1 2 We also have 2 + 3 + 3 for 0 so in fact we get This means that 1 + p +2 t (mod p +3 ). (mod p +3 ). (1 + p) p+1 = 1 + p +2 t + p +3 m = 1 + p +2 (t + pm) for some m Z. Since gcd(t, p) = 1, we also have gcd(t + pm, p) = 1 and hence the inductive step is correct. For e 1, the above computation shows that (1 + p) pe 1 1 (mod p e ) and that there exists t Z relatively prime to p with (1 + p) pe 2 1 + p e 1 t 1 (mod p e ). Thus the order of 1 + p is indeed p e 1 in (Z/p e Z). Corollary 3.3. Let p be an odd prime. Then the group (Z/p e Z) is cyclic. Proof. Proposition 3.1 shows that (Z/p e Z) contains an element of order p 1 while Proposition 3.2 shows that (Z/p e Z) contains an element of order p e 1. By Proposition 2.1 we see that (Z/p e Z) contains an element of order lcm(p 1, p e 1 ) = (p 1)p e 1 = ϕ(p e ) and hence (Z/p e Z) is cyclic. 5
4 The case p = 2 The above proof cannot wor for p = 2. The group is not cyclic since (Z/8Z) = {1, 3, 5, 7} 1 1 2 3 2 5 2 7 2 (mod 8). Before proving what is correct, let s try to understand what fails in the proof. The fact that (Z/2 e Z) contains an element of order 2 1 = 1 is trivial. Thus something must fail with the order of 1 + 2. We have (1 + 2) 20 = 1 + 2 1 = 1 + 2 1 1 (1 + 2) 21 = 1 + 4 + 4 = 1 + 2 3 1. Strangely enough, the only thing that goes wrong in the induction is the claim from above that ( ) p p(p 1) = 2 2 is a multiple of p. This is true if p is odd, but false for p = 2. We will need a new idea here. We will prove that 5 always has order 2 e 2 in (Z/2 e Z). The extra power of 2 in 5 = 1 + 2 2 will be useful for fixing the proof. Proposition 4.1. The element 5 = 1 + 2 2 has order p e 2 in (Z/p e Z) whenever e 2. Proof. We prove by induction the stronger statement that for 0, (1 + 2 2 ) 2 = 1 + 2 +2 t where t Z is odd. This is clearly true for = 0 and t = 1 so assume it is true for some 0. Then (1 + 2 2 ) 2+1 = (1 + 2 +2 t) 2 = 1 + 2 2 +2 t + 2 2+4 t 2 = 1 + 2 +3 (t + 2 +1 t 2 ) and t + 2 +1 t 2 is odd since t is odd. This proves the result by induction, and hence 5 2e 2 1 (mod 2 e ) 6
yet for some odd t Z, 5 2e 3 = 1 + 2 e 1 t 1 (mod 2 e ). This proves that 5 has order 2 e 2 is (Z/2 e Z). This shows that 5 (Z/2 e Z) is a subgroup of index 2 since ϕ(2 e ) = 2 e 1. We will prove that 1 5. To do this, notice that the order of 1 in (Z/2 e Z) is 2. Any cyclic group has at most one element of order 2, so all we need to do is to find an element of order 2 in 5 and chec that it is not 1. Proposition 4.2. We have 1 5 in (Z/2 e Z) whenever e 2. Proof. This is obvious if e = 2. Thus we may assume that e 3. We proved in Proposition 4.1 that the order of 5 is 2 e 2. Thus 5 2e 3 has order 2 in 5. Since 1 also has multiplicative order 2, if 1 5 then we must have But then where t is some odd integer. Thus 5 2e 3 1 (mod 2 e ). 1 5 2e 3 1 + 2 e 1 t (mod 2 e ) 2 2 e 1 (mod 2 e ) which is absurd since 2 e 1 divides 2 e yet not 2 as long as e 3. Corollary 4.3. For e 2 the group (Z/2 e Z) is isomorphic to the direct product of a cyclic group of order 2 and a cyclic group of order 2 e 2. Proof. Consider subgroups A = 5 (Z/2 e Z) B = 1 (Z/2 e Z). Proposition 4.1 and Proposition 4.2 guarantee that A = 2 e 2, B = 2 and A B = 1. Since (Z/2 e Z) is abelian, we have Of course and therefore as claimed. AB = AB = A B. A B A B = (2e 2 ) 2 = 2 e 1 = (Z/2 e Z) (Z/2 e Z) = A B 7