Suppose \(G = \langle g\rangle \) is a cyclic group and \(H \leq G\). If \(H = \{ e\} \), where \(e\) denotes the identity element of \(G\), then \(H = \langle e\rangle \). So \(H\) is cylic. So, from now on, assume that \(H \neq \{ e\} \).

Set \(L := \{ n\in \mathbb {Z} : g^n \in H\} \) and set \(L^+ := L^+\cap \mathbb {P}\).

I claim that \(L^+\) is nonempty. To see this, using the assumption that \(H \neq \{ e\} \), pick \(h\in H\setminus \{ e\} \). Since \(G = \langle g\rangle \), we can find \(n\in \mathbb {Z}\) such that \(h = g^n\). Moreover, \(n\neq 0\), since \(g^0 = e\) by definition. So either \(n {\gt} 0\) or \(-n {\lt} 0\). If \(n {\gt} 0\), then \(n\in L^+\). Otherwise, \(g^{-n} = h^{-1}\in H\). So \(-n \in L^+\).

Now, since \(L^+\) is nonempty, it has a smallest element \(d\in L^+\) by the well-ordered property of \(\mathbb {P}\). I claim that \(H = \langle g^d\rangle \). It’s clear that \(\langle g^d \rangle \leq H\) since \(g^d\in H\). So we need to show that \(H \leq \langle g^d\rangle \). To see this, pick \(h\in H\) and write \(h = g^n\) for some \(n\in \mathbb {Z}\). Then, using the division algorithm, write \(n = qd + r\) for integers \(q\) and \(r\) with \(0\leq r {\lt} d\). We get that \(h = g^n = g^{qd + r} = (g^{d})^q g^r\). So, as \(h, g^d\in H\), \(g^r = h(g^{d})^{-1}\in H\) as well. By the minimality of \(d\) and the fact that \(0\leq r {\lt}d\), this then implies that \(r = 0\). So \(h = (g^d)^q\in \langle g^d\rangle \). Thus, as \(h\) was arbitrary, \(H = \langle g^d\rangle \) proving the theorem.

Obvious from the theorem as \(\mathbb {Z}\) is cyclic and \(d\mathbb {Z} = (-d)\mathbb {Z}\) for \(d\in \mathbb {Z}\).

Suppose \(G = \langle g\rangle \) is cyclic and \(x,y\in G\). We can find integers \(n,m\) such that \(x = g^n\) and \(y = g^m\). Then \(xy = g^ng^m = g^{n+m} = g^m g^n = yx\).

The symmetric group \(S_3\) is not abelian.