设 G = < a > G=<a> G=<a> Is a cyclic group
Theorem 1: If GGG is an infinite cyclic group, thenGGThe generator of G is onlyaaa suma − 1 a ^ {-1}a− 1
Theorem 2:IfGGG isnnfinite cyclic group of order n , thenGGThe generators of G have a total ofφ (n) \varphi(n)φ ( n ) , they areak (1 ⩽ k ⩽ n, (k, n) = 1 a^k(1\leqslant k\leqslant n,(k,n)=1ak(1⩽k⩽n,(k,n)=1).
Proof of Theorem 1: Obviously ⟨a − 1⟩ = ⟨a⟩ = G. \Langle a^{-1}\rang=\langle a\rangle=G.⟨a−1⟩=⟨a⟩=G . On the other hand, since the[G: ⟨an⟩] =[G:⟨an⟩]=∣ n ∣ , we know that⟨an⟩ = G ⟺ ∣ n ∣ = 1 ⟺ n = ± 1 \lang a^n\rang=G\iff|n|=1\iff n=\pm 1⟨an⟩=G⟺∣n∣=1⟺n=±1
Proof of Theorem 2: Since aaa of the order ofnnn , soaka^kaThe order of k isn / (k, n) n/(k,n)n/(k,n ) . So⟨ak⟩ = G ⟺ ak \lang a^k\rang=G\iff a^k⟨ak⟩=G⟺aThe order of k isnnn ⟺ ( k , n ) = 1 \iff (k,n)=1 ⟺(k,n)=1