# ALGEBRA I (LECTURE NOTES 2017/2018) LECTURE 9 - CYCLIC GROUPS AND EULER S FUNCTION

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1 ALGEBRA I (LECTURE NOTES 2017/2018) LECTURE 9 - CYCLIC GROUPS AND EULER S FUNCTION PAVEL RŮŽIČKA 9.1. Congruence modulo n. Let us have a closer look at a particular example of a congruence relation on the group Z = (Z, +) of all integers with the operation of addition. Since the group Z is commutative, all subgroups of Z are normal (cf. Remark 6.2). For a positive integer n there is the subgroup nz with the universe nz := {na a Z} = {b Z n b} of the group Z. In fact, these are the only subgroups of Z: Lemma 9.1. Let A be a non-trivial subgroup of Z. Then A = nz where n is the smallest positive integer from A. Proof. Let A = (A, ) be a non-trivial subgroup of Z. Since A is nontrivial, it contains a positive integer, indeed, if s < 0 belongs to A, then 0 < s A as well. Let n be the smallest positive integer in A. Since nz is clearly the least subgroup of Z containing n, we have that nz A. Let s A. Dividing s by n with remainder, we find integers t, r such that s = n t = r and 0 r < n. From r = s n t A we infer that r = 0, since otherwise it will violate the choice of n. Therefore n s, and so s nz. We conclude that A nz, and so the two subgroups are equal. Adding the trivial subgroup to the picture we get that Corollary 9.2. All subgroups of the group Z are of the form nz for some non-negative integer n. We say integer s is congruent with an integer t modulo n, and write s t (mod n), if s is congruent with t modulo nz, that is, if s nz t. By the definition of the congruence relation modulo a normal subgroup in Subsection 7.5 (and the commutativity of the group Z), we have that (9.1) s t (mod n) if and only if n t s. Date: December 4,

2 2 PAVEL RŮŽIČKA It follows from Lemma 7.23 that Corollary 9.3. Let n be an integer. The following properties hold true. (i) If s 1 t 1 (mod n) and s 2 t 2 (mod n), then s 1 + s 2 t 1 + t 2 (mod n), for all s 1, s 2, t 1, t 2 Z. (ii) If s t (mod n), then s t (mod n), for all s, t Z. Exercise 9.1. Prove Corollary 9.3 readily from the definition of the congruence modulo n. Let us denote by gcd(s, t) and lcm(s, t) respectively the greatest common (non-negative) divisor and the least common (non-negative) multiple of integers s, t. The next exercises cover some additional properties of congruences modulo positive integers. Exercise 9.2. Let n be a positive integer. Prove that (i) if s 1 t 1 (mod n) and s 2 t 2 (mod n), then s 1 s 2 t 1 t 2 (mod n), for all s 1, s 2, t 1, t 2 Z. (ii) if s t (mod n), then s k t k (mod n), for all s, t Z and all k N. Exercise 9.3. Prove that (i) if su tu (mod n) and gcd(u, n) = 1, then s t (mod n), for all s, t, u Z and n N. (ii) if s t (mod m i ) for all m 1,..., m k, then s t (mod lcm(m 1,..., m k )), for all s, t Z and m 1,..., m k N Transversals. Let G be a group and H a subgroup of the group G. A left (respectively right) transversal for H is a set picking one element from each left (respectively right) coset of H. Clearly, the size of a left (respectively right) transversal, say L (respectively R), equals to the size of the set of all left (respectively right) cosets of H. That is L = R = [G : H]. If N is a normal subgroup of the group G, then left and right transversals for N coincide. In this case we will talk about transversals for N. Assume that we have a group G, a normal subgroup N of G, and a transversal T for N, often containing the unit of G. If we have a

3 nice algorithm that for a given a, b T computes c T such that a b N = c N, we can view the elements of the factor group G/N as elements of the transversal T. The group operation of G/N will then be determined by our algorithm. The set Z n := {0, 1,..., n 1} is a transversal for the subgroup nz in the additive group Z of all integers. Given an integer s, we divide s by n with reminder. We denote the reminder by s mod n. This allows us to define a binary operation on the set Z n,denoted by + n, as follows: s + n t := (s + t) mod n Z n for all s, t Z n. We will call the operation + n the addition modulo n. As discussed above Z n = (Z n, + n ) is a group. The homomorphism π n : Z Z n, given by s s mod n, maps Z onto Z n. It is straightforward that ker π n = nz, and so there is a (unique) isomorphism µ n : Z/nZ Z n such that π n = µ n π Z/nZ due to The homomorphism theorem. The isomorphism µ n is given by the correspondence s nz s mod n Cyclic groups. The order of a finite group is the number of its elements while the order of an infinite group is set to be. A group C is cyclic if it is generated by a single element, say g. We will use the notation g for the cyclic group generated by g. Observe that all elements of g are powers of g, and the map ε g : Z g given by s g s is a group homomorphism onto g. According to The first isomorphism theorem g Z/ ker ε g. Applying Lemma 9.1 it follows that either ker ε g = 0 and g Z or ker ε g = nz for some positive integer n and g Z/nZ Z n. In the latter case, n is the order of g as well as the order of g. We showed that Theorem 9.4. Up to isomorphism the cyclic groups are Z and Z n, n N. The group Z is of an infinite order while the order of Z n is n. In particular, a cyclic group is determined by its order up to isomorphism. Lemma 9.5. Let φ: G H be a group homomorphism and K a subgroup of the group H. Then is a subgroup of G. φ 1 (K) := {g G φ(g) K} Proof. Since φ(u G ) = u H K, we have that u G φ 1 (K). In particular, φ 1 (K) is non-empty. If g, h φ 1 (K), then φ(g h 1 ) = φ(g) φ(h) 1 K, hence g h 1 φ 1 (K). Lemma 9.6. Every factor-group and every subgroup of a cyclic group is cyclic. 3

4 4 PAVEL RŮŽIČKA Proof. Let C is a cyclic group generated by g. Then a factor group of C is generated by the coset of g, in particular it is cyclic. According to Lemma 9.1 non-trivial subgroups of Z are isomorphic to Z. In particular, every subgroup of Z is cyclic. If D is a subgroup of C, then ε 1 g (D) is a subgroup Z and D is its homomorphic image. We conclude that D is cyclic. Let m, n be positive integers such that m n. Then nz mz and the group mz/nz is cyclic due to Lemma 9.6. Observe that ker ε 1+m nz = (n/m)z, hence (9.2) mz/nz Z n/m. Lemma 9.7. If m divides n, there is a unique subgroup of Z n of order m. Proof. It follows from (9.2) that ε 1 ((n/m)z) is a subgroup of Z n of order m. On the other hand, if D is a subgroup of Z n of order m, then ε 1 1 (D) = (n/m)z, again due to (9.2). It follows that the subgroup of Z n of order m is unique Orders of elements. Let G = (G, ) be a group and g G. We set g 0 := u G, g n := g g }{{} n times Remark 9.8. Observe that (i) g s t = (g s ) t, (ii) g s+t = g s g t, for all g G, and all s, t Z. and g n := g 1 g }{{ 1, for all n N. } n times An order of an element g of a group G, denote by o(g), is the least n > 0 such that g n = u G. If no such n exists, we put o(g) :=. In the first case we say that g has a finite order, in the latter we say that g has an infinite order. Lemma 9.9. The order of an element g of a finite group G divides the order of the group. Proof. The order, o(g), of an element g equals to the order of the cyclic group g generated by g. The order of the subgroup g divides the order of G, due to the Lagrange theorem. Lemma Let G = (G, ) be a group and g G. Then (i) o(g) = if and only if g s g t for all pairs of distinct integers s, t.

5 (ii) If the element g is of a finite order, then g s = g t if and only if s t (mod o(g)), for all s, t Z. In particular, g s = u G if and only if o(g) s. Proof. Since g s+t = g s g t, for all s, t Z, the map ε: Z G given by s g s is a group homomorphism. It follows from the definition and Corollary 7.11 that o(g) = if and only if ker ε = 0 if and only if φ is one-to-one. This settles (i) and implies that the element g has a finite order if and only if the kernel of ε is non-trivial. If this is the case then ker ε = nz wher 0 < n = o(g), due to Lemma 9.1. It follows that g s = g t if and only t s ker ε if and only if t s (mod n). Since u G = g 0 we conclude that g s = u G if and only if s 0 (mod n). This is exactly when o(g) s, and so we have proved (ii). Recall that integers s and t are said to be relatively prime provided that gcd(s, t) = 1. Lemma Let G = (G, ) be a group and f, g G elements of a finite order such that f g = g f. Then the following holds true: (i) o(f g) lcm(o(f), o(g)). (ii) if gcd(o(f), o(g)) = 1, then o(f g) = o(f) o(g). Proof. (i) Put m = lcm(o(f), o(g)) and observe that f m = g m = u G, indeed, both o(f) m and o(g) m hold true. Since the elements f and g commute, we get that (f g) m = f m g m = u G. It follows that o(f g) lcm(o(f), o(g)) due to Lemma (ii) Put n = o(f g). It follows from (i) that n o(f) o(g). Since f and g commute we have that u G = (f g) n o(g) = f n o(g) g n o(g) = f n o(g) (g o(g) ) n = f n o(g). It follows from Lemma 9.10 that o(f) n o(g) and since o(f) and o(g) are relatively prime, we get that o(f) n. Similarly we prove that o(g) n and since gcd(o(f), o(g)) = 1, we conclude that o(f) o(g) n. Therefore n = o(f) o(g). Corollary Let G = (G, ) be a group and f, g G commuting elements of a finite order. Putting m = gcd(o(f), o(g)), we get that o(f m g) = o(f g m ) = By induction we prove that o(f) o(g) gcd(o(f), o(g)) = lcm(o(f), o(g)). Corollary Let g 1,..., g k be commuting elements of a finite order of a group G. (i) Then o(g 1 g k ) lcm(o(g 1 ),..., o(g k )). 5

6 6 PAVEL RŮŽIČKA (ii) If o(g 1 ),..., o(g n ) are pairwise relatively prime, then o(g 1 g k ) = o(g 1 ) o(g k ). (iii) There are m 1,..., m k 1 N such that o(g m 1 1 g m k 1 k 1 g k ) = lcm(o(g 1 ),..., o(g k )). Exercise 9.4. Let π = γ 1 γ k be a decomposition of a permutation π S n into the product of independent cycles. Prove that o(π) = lcm( γ 1,..., γ k ). Corollary Let F = (F, ) be a finite abelian group and g F an element of the maximum order in A. Then o(f) o(g) for all f F. Proof. According to Corollary 9.13 (iii), there is g F such that o(g) = lcm({o(f) f F }). Theorem Every finite subgroup of the multiplicative group F = (F \ {0}, ) of a field F is cyclic. Proof. Let G be a finite subgroup of F. Let n be maximum order of an element of G. It follows from Corollary 9.14 that o(g) n for all g G, hence every element of G is a root of the polynomial x n 1. This polynomial has at most n-distinct roots, hence G n. On the other hand n G as it follows from Lemma 9.9. We conclude that n = G. Therefore the group G is cyclic. Example There is no bound of o(f g) by o(f) and o(g) in general. For example let n N and π := (1, 2n) (2, 2n 1) (3, 2n 2) (n, n + 1), σ := (2, 2n) (3, 2n 1) (4, 2n 2) (n, n + 2) be permutations from S 2n. Since both π and σ are products of independent transpositions o(π) = o(σ) = 2. Computing that σ π := (1, 2, 3,..., 2n) is a 2n-cycle, we get that o(σ π) = 2n. Exercise 9.5. Can you guess the product π σ without computing it? Exercise 9.6. Let π and σ be a as in Example Put ρ := (1, n + 1) σ and compute that o(π) = o(ρ) = 2 while o(ρ π) = n.

7 9.5. Euler s function. The cyclic group Z of an infinite order has exactly two generators 1 and 1. A cyclic group of a finite order n is isomorphic to Z n = (Z n = {0, 1,..., n 1}, + n ). The following are equivalent for an element s Z n : (i) s is a generator of Z n, (ii) o(s) = n, (iii) ks 0 (mod n) for all k = 1, 2,..., n 1, (iv) gcd(s, n) = 1. For a positive integer n we denote by Z n the set of all generators of the group Z n, i.e, (9.3) Z n := {s {1,..., n} gcd(s, n) = 1}. The Euler s function is a map φ: N N which assigns to a positive integer n the number of generators of Z n, i.e, φ(n) = Z n, for all n N. Lemma Let p be a prime and m N. Then φ(p m ) = p m p m 1. Proof. Since p is a prime, we have that {s {1, 2,..., p m } p divides s} = {pt t {1,..., p m 1 }}. Therefore the set {1, 2,..., p m } contains exactly p m 1 numbers divisible by p. The rest are elements relatively prime to p, thus φ(p m ) = p m p m 1. The cartesian product G 1 G n of groups G 1,..., G n consists of all tuples g 1,..., g n such that g i G i for all i {1, 2,..., n}. Elements of G 1 G n are multiplied coordinate-wise. Lemma Let n 1,..., n k be positive integers. If they are pairwise relatively prime, then φ(n 1 n k ) = φ(n 1 ) φ(n k ). Proof. Let s 1,..., s k be an element Z n1 Z nk. Since o(s i ) n i, for all i = 1,..., k, the orders o(s 1 ),..., o(s k ) are relatively prime. We get that o( s 1,..., s k ) = o(s 1 ) o(s k ), due to Lemma 9.13 (i). It follows that s 1,..., s k is a generator of Z n1 Z nk if and only if s i generates Z ni for all i {1, 2,..., k}. We conclude that the cartesian product Z n1 Z nk is cyclic of order n 1 n k and φ(n 1 n k ) = φ(n 1 ) φ(n k ). Exercise 9.7. Prove that Z n1 Z nk is cyclic if and only if n 1,, n k are pairwise relatively prime. 7

8 8 PAVEL RŮŽIČKA Theorem Let n = p m 1 1 p m k k be a decomposition of a positive integer n into the product of primes. Then φ(n) = (p m 1 1 p m ) (p m k k p m k 1 k ) = n(1 1 ) (1 1 ). p 1 p k Proof. We have that φ(n) = φ(p m p m k k ) = φ(pm 1 Lemma 9.18 and φ(p m i i Lemma ) = p m i i p m i 1 i 1 ) φ(p m k k ) due to, for all i = 1,..., k, due to Theorem For every positive integer n the equality (9.4) n = m n φ(m) holds true. Proof. Observe that an element g of a group G is a generator of a unique subgroup of G, namely the cyclic subgroup g consisting of all powers of g. The cyclic group Z n has n elements, a unique subgroup of order m for each m n, due to Lemma 9.7, and the subgroup of order m has exactly φ(m) generators. Equality (9.4) follows. The multiplication modulo a positive integer n is given by s n t = s t mod n, is a binary operation on Z n. It follows from Exercise 9.2 that the set Z n := {s Z n s 1 (mod n)} together with the operation n form a group. We will denote the group by Z n. Theorem 9.21 (Euler s theorem). Let n be a positive integer. Then (9.5) s φ(n) 1 (mod n), for all integers s co-prime to n. Proof. Let s be an integer co-prime to n. Then s t (mod n) for some t Z n. The order of t in Z n divides φ(n) = the order of Z n, due to Lemma 9.9. It follows from Lemma 9.10 that t n n t = 1, }{{} φ(n) times hence t φ(n) 1 (mod n). Since s φ(n) t φ(n) (mod n), due to Exercise 9.2, we conclude that (9.5) holds true. Corollary 9.22 (Fermat s theorem). Let p be a prime. If p s, then s p 1 1 (mod p).

9 Proof. Since p is prime, an integer s is co-prime to p if and only if p s. Since φ(p) = p 1 for every prime p, Fermat s theorem follows readily form Euler s theorem. Lemma 9.23 (Wilson s theorem). Let 1 < q be an integer. Then q (q 1)! + 1 if and only if q is a prime. Proof. ( ) If q is not prime, then clearly 1 < gcd(q, (q 1)!), and so q (q 1)! + 1. ( ) Suppose that q is a prime number. Then q s 2 1 = (s + 1)(s 1) if and only if q s + 1 or q s + 1. It follows that the only elements of the group Z q that are equal to their inverses are 1 and q 1. Consequently, we can pair the remaining elements of Z q, namely 2,, q 2, so that the members of every pair are inverse to each other. We conclude that 2 (q 2) 1 (mod q), hence (q 1)! (q 1) 1 (mod q), whence q (q 1)!

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