Since \(\Z\) and each \(\Z_n\) are abelian, every cyclic group is abelian. But, for instance, \(\R\) is an abelian noncyclic group (it can't be cyclic because it is uncountable).

Let \(x,y\in \langle a\rangle\text{.}\) Then \(x=a^i\text{,}\) \(y=a^j\) for some \(i,j\in \Z\text{.}\) So \(xy=a^ia^j=a^{i+j}\in \langle a\rangle\text{.}\) Next, \(e_G=a^0\in \langle a \rangle\text{.}\) Finally, \((a^i)^{-1}=a^{-i}\in \langle a \rangle\text{.}\) So \(\langle a\rangle\) is a subgroup of \(G\text{.}\)

Case 1. There is no positive integer \(k\) with \(a^k=e\text{.}\) In this case, we claim \(\langle a\rangle \simeq \Z\text{.}\) Indeed: Let \(\phi:\Z\to \langle a\rangle\) be defined by \(\phi(i)=a^i\text{.}\) Clearly, \(\phi\) is a homomorphism that is onto. So to show that \(\phi\) is an isomorphism, it suffices to show that \(\phi\) is one-to-one. Let \(i,j\in \Z\) with \(\phi(i)=\phi(j)\text{.}\) Then \(a^i=a^j\text{.}\) Without loss of generality, we may assume \(i\geq j\text{.}\) Then, multiplying both sides of the equation by \(a^{-j}\) (on the right or on the left) we obtain \(a^{i-j}=e\text{.}\) Since \(i-j\in \N\) and there is no positive integer \(k\) with \(a^k=e\text{,}\) we must have \(i-j=0\text{,}\) so \(i=j\text{.}\) Thus, \(\phi\) is an isomorphism from \(\Z\) to \(\langle a\rangle\text{.}\)

Case 2. There is a positive integer \(k\) such that \(a^k=e\text{.}\) In this case, let \(n\) be the least such positive integer. Define \(\phi:\Z_n\to \langle a\rangle\) be defined by \(\phi(i)=a^i\text{.}\) We claim that \(\phi\) is an isomorphism.

First, let \(i,j\in \Z\text{.}\) We want to show that \(\phi(i+_nj)=\phi(i)\phi(j)\text{,}\) that is, that \(a^{i+_nj}=a^ia^j\text{.}\) Since \(i+j\) is congruent to \(i+_nj\) modulo \(n\text{,}\) \(a^ia^j=a^{i+j}=a^{i+_nj}\text{,}\) by the Clock Lemma. So \(\phi\) is a homomorphism. Next we show that \(\phi\) is one-to-one. Let \(i,j\in \Z_n\) with \(\phi(i)=\phi(j)\text{,}\) so \(a^i=a^j\text{.}\) Then, by the Clock Lemma, \(i\) and \(j\) are congruent modulo \(n\text{.}\) Since they're both in \(\Z_n\text{,}\) they must be equal, so \(\phi\) is one-to-one. Finally, we show that \(\phi\) is onto. let \(x\in \langle a\rangle\text{.}\) Then \(x=a^i\) for some \(i\in \Z\text{.}\) Letting \(r\) be the remainder when we divide \(i\) by \(n\text{,}\) we have \(r\in \Z_n\) with \(i\) congruent to \(r\) modulo \(n\text{;}\) so, again using the Clock Lemma, \(x=a^i=a^r\text{.}\) Since \(r\in \Z_n\text{,}\) \(x=\phi(r)\text{.}\) So \(\phi\) is onto.

Thus, \(\phi\) is an isomorphism, and so in this case we have \(\Z_n \simeq \langle a \rangle\text{.}\)

This follows from the last statement of the Clock Lemma.

First, assume \(o(a)=n\lt \infty\text{.}\) Then

so \(o(a^{-1})\leq n=o(a)\text{.}\) Using the same argument, we have \(n=o(a)\leq o(a^{-1})\text{.}\) Since \(o(a^{-1})\leq n\) and \(n\leq o(a^{-1})\text{,}\) \(o(a^{-1})=n=o(a)\text{.}\)

On the other hand, assume \(o(a)= \infty\text{.}\) Then \(a^{-1}\) must also have infinite order, since if it had finite order \(m\text{,}\) \(a\) would, by the above argument, have order less than or equal to \(m\text{.}\)

Let \(k\in \Z^+\text{.}\) By definition of \(c\) and \(d\) and by Theorem 5.1.12, we have \(k(a,b)=(ka,kb)=(0,0)\) if and only if \(c\) and \(d\) divide \(k\text{.}\) Since \(\gcd(c,d)=1\text{,}\) the least positive integer divisble by both \(c\) and \(d\) is \(\lcm(c,d)\text{.}\) Thus, \(o((a,b))=\lcm(c,d)\text{,}\) as desired.

Let \(G\) be cyclic. Suppose \(|G|=\infty\text{.}\) Then there is an isomorphism \(\phi: \Z\to G\text{.}\) Note that \(\Z=\langle 1\rangle\text{.}\) So

A similar argument shows that there exists \(a\in G\) such that \(G=\langle a\rangle\) when \(|G|\lt \infty\text{.}\)

This follows from the fact that \(\phi(\langle a\rangle )=\langle \phi(a)\rangle\text{.}\)