= Solution
\b[Always true.] Suppose $G=\langle g\rangle$. If $H$ is nontrivial, let $d$ be the least positive integer with $g^d\in H$. For any $g^k\in H$, divide $k=qd+r$ with $0\le r<d$. Since $g^r=g^k(g^d)^{-q}\in H$, minimality forces $r=0$. Hence $H=\langle g^d\rangle$ is a <cyclic group>. The trivial <subgroup> is cyclic as well. Every <coset> in $G/H$ is $g^kH=(gH)^k$, so the <quotient group> is generated by $gH$. This proves the <subgroups and quotients of a cyclic group> assertion.
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