Cyclic Group • Definition ○ A group G is cyclic if ∃g∈G s.t. ⟨g⟩=G • Note ○ A finite group G of order n is cyclic iff ∃g∈G s.t. |g|=n • Example 1: Z is cyclic ○ Z=⟨1⟩ ○ Z=⟨−1⟩ • Example 2: Z\nZ is cyclic ○ If (a,n)=1, Z\nZ=⟨a ̅ ⟩ ○ (We will prove this later) • Example 3: S_3 is not cyclic ○ Note: If (a_1,…,a_t )∈S_n is a t-cycle, then |(a_1,…,a_t )|=t ○ S_3={(1),(1 2),(1 3),(2 3),(1 2 3),(1 3 2)} ○ Every element in S_3 have order 1,2, or 3 ○ So S_3 cannot be cyclic Proposition 18 • Let G be a cyclic group • If |G|=∞, then G≅Z ○ Choose g∈G s.t. G=⟨g⟩ ○ Define a map f:Z→G by n↦g^n ○ Check homomorphism § If n_1,n_2∈Z § then f(n_1+n_2 )=g^(n_1+n_2 )=g^(n_1 ) g^(n_2 )=f(n_1 )f(n_2 ) § Thus, f is a homomorphism ○ Check surjectivity § Surjectivity is clear ○ Check injectivity § Suppose f(n_1 )=f(n_2 ) § Then g^(n_1 )=g^(n_2 ) § Without loss of generality, assume n_1≥n_2 § Then g^(n_1−n_2 )=1 § Since |g|=∞ § n_1−n_2=0 § i.e. n_1=n_2 § Thus f is injective • If |G|=n ∞, then G≅Z\nZ ○ Choose g∈G s.t. G=⟨g⟩ ○ Define a map f:Z\nZ→G by a ̅↦g^a ○ Check well-definedness § We need to check that f is well-defined. § That is we must show that if a ̅=b ̅ in Z\nZ, then f(a ̅ )=f(b ̅ ) § Let a,b∈Z, suppose a ̅=b ̅ in Z\nZ § Choose q∈Z s,t, nq=a−b § f(a ̅ )=ga=g(nq+b)=g^nq gb=gb=f(b ̅ ) § Thus, f is well-defined ○ Check homomorphism § f(a ̅+b ̅ )=g(a+b)=ga g^b=f(a ̅ )f(b ̅ ) § Thus, f is a homomorphism ○ Check surjectivity § Surjectivity is clear ○ Check injectivity § If f(a ̅ )=f(b ̅ ) § ga=gb § g^(a−b)=1 § ├ |g|┤|├ (a−b)┤ § n|(a−b) § a ̅=b ̅ § Thus f is injective Least Common Multiple • Definition ○ Let a,b∈Z where one of a,b is nonzero. ○ A least common multiple of a and b is a positive integer m s.t. § a|m and b|m § If a|m′ and b|m′, then m|m^′ ○ We denote the lcm of a and b by [a,b] ○ Define [0,0]≔0 • Uniqueness ○ Similar to the proof of uniqueness of gcd • Existence: If a,b∈Z, and one of a,b is nonzero, then ab/((a,b) ) is the lcm of a,b ○ ab/((a,b) ) is a multiple of a and b ○ Suppose m^′∈Z s.t. a|m′ and b|m′ ○ We must show ├ ab/((a,b) )┤|├ m′┤ ○ Choose q,q^′∈Z s,t, aq=m′ and bq^′=m′ ○ Choose x,y∈Z s.t. ax+by=(a,b) ○ m^′ (a,b)=m^′ (ax+by)=m^′ ax+m^′ by=bq^′ ax+aqby=ab(q^′ x+qy) ○ Thus ab|(m^′ (a,b)) ○ i.e. ab/((a,b) )=m^′ Proposition 19 • Statement ○ If G=⟨g⟩ is cyclic, and |G|=n ∞, then |g^a |=n/((a,n) ) • Proof ○ Let a∈Z ○ If a=0, this is clear ○ So, assume a≠0 ○ (g^a )^(n/((a,n) ))=g^(an/((a,n) ))=g^[a,n] =g^kn for some integer k ○ Thus, (g^a )^(n/((a,n) ))=(g^n )^k=1 since n=|g| ○ Suppose t∈Z( 0) and (g^a )^t=1 ○ By HW3, g^at=1⇒n|at ○ Thus, at is a common multiple of n and a ○ ├ [a,n]┤|├ at┤ ○ ⇒├ an/((a,n) )┤|├ at┤ ○ ⇒├ n/((a,n) )┤|├ t┤ ○ In particular, n/((a,b) )≤t Theorem 20 • Let G=⟨g⟩ be a cyclic group • Statement (1) ○ Every subgroup of G is cyclic ○ More precisely, if H≤G, then either H={1} or ○ H=⟨g^d ⟩, where d is the smallest positive integer s.t. g^d∈H • Statement (2) ○ If G is finite, then for all positive integers a dividing n ○ ∃! subgroup H≤G of order a. ○ Moreover, this subgroup is ⟨g^d ⟩, where d=n/a
