Solution. An element \(k\) generates \(\mathbb {Z}_n\) exactly when \(\gcd (k,n)=1\), because the subgroup generated by \(k\) has order \(n/\gcd (k,n)\) and this equals \(n\) only when the gcd is \(1\).

For \(n=12\) the units are \[1,\ 5,\ 7,\ 11 ,\] four of them, which is \(\varphi (12)=4\). The general rule is that \(\mathbb {Z}_n\) has exactly \(\varphi (n)\) generators.

Note that \(\mathbb {Z}_{12}\) is cyclic and therefore abelian, yet most of its elements do not generate it. Being cyclic is a statement about the existence of one generator, not about all elements.