= Solution
A <group> $G$ is <abelian group>[abelian] when $xy=yx$ for every $x,y\in G$. It is <cyclic group>[cyclic] when some <generator of a group> $g$ satisfies
$$
G=\langle g\rangle=\{g^n:n\in\mathbb Z\}.
$$
If $x=g^r$ and $y=g^s$ belong to a cyclic group, then
$$
xy=g^{r+s}=g^{s+r}=yx,
$$
so every cyclic group is abelian.
The <Klein four-group> $C_2\times C_2$ is abelian, but every nonidentity element has order two, so no element generates all four elements. Thus an abelian group need not be cyclic.
The condition on proper <subgroups> does not force $G$ to be abelian. The <quaternion group>
$$
Q_8=\{\pm1,\pm i,\pm j,\pm k\}
$$
is nonabelian because $ij=k$ whereas $ji=-k$. Its proper subgroups are the trivial subgroup, $\{\pm1\}$, and the three cyclic subgroups
$$
\langle i\rangle,\qquad\langle j\rangle,\qquad\langle k\rangle,
$$
each of order four. They are all cyclic, so
$$
\boxed{\text{the answer is no}}.
$$
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