Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.12
Prove that the following groups are not cyclic: (1) $Z_2 \times Z_2$, (2) $Z_2 \times \mathbb{Z}$, and (3) $\mathbb{Z} \times \mathbb{Z}$. (We suppose $Z_2 = \langle x \rangle$.)
Solution: First we prove a lemma.
Lemma: Let $G$ and $H$ be groups. If $G \times H$ = $\langle (g,h) \rangle$ is cyclic, then $G = \langle g \rangle $and $H = \langle h \rangle$.
Proof: Let $x \in G$ and $y \in H$. Then $(x,y) = (g,h)^n = (g^n,h^n)$ for some integer $n$. Thus $x = g^n$ and $y = h^n$, so the conclusion follows. $\square$
(1) If $Z_2 \times Z_2 = \langle (a,b) \rangle$, then by the lemma, $a = b = x$. But then $$\langle (x,x) \rangle = \{ (1,1), (x,x) \},$$ a contradiction.
(2) If $Z_2 \times \mathbb{Z} = \langle (a,b) \rangle$, then $a = x$ and either $b = 1$ or $b = -1$.
Suppose $b = 1$; then $(x,2) = (x,1)^n = (x^n,n)$ for some $n$; then $n = 2$ and so $x^2 = x$, a contradiction.
If $b = -1$, then $$(x,2) = (x,-1)^n = (x^{-n}, -n)$$ for some $n$, so $n = -2$ and thus $x = x^2$, a contradiction. So $Z_2 \times \mathbb{Z}$ is not cyclic.
(3) If $\mathbb{Z} \times \mathbb{Z} = \langle (a,b) \rangle$ is cyclic, then by the lemma we have $a = \pm 1$ and $b = \pm 1$. Note, however, that $$(\pm 1, \pm 1)^n = (\pm n, \pm n),$$ so that $(\pm 1, \pm 1)$ does not generate all of $\mathbb{Z} \times \mathbb{Z}$.
