When $G$ is a cyclic group, any subgroup $H$ of it is also cyclic and hence the quotient group $G/H$ is also cyclic. Moreover, if $a\in G\setminus H$ be a generator of $G$ then $\langle aH\rangle$ is a generator of $G/H$. While I was studying the proof I, I was curious to know the converse part in a special case.
Suppose $G$ is a finite cyclic group and $G/H$ is cyclic with $aH$ as one of its generators where $a\in G\setminus H$. Can we say $a$ is a generator of $G$ as well?
I know that there are plenty non-cyclic groups possessing cyclic quotient groups(for example $S_3$ is non-cyclic and $S_3/A_3$ is cyclic), but if it is stated beforehand that $G$ is cyclic, I don't know how to proceed.
I was hoping to find a counter-example but in vain. Neither I was convinced that the statement is false. Any help please.
Let $G$ be cyclic of order $6$, written multiplicatively and generated by $x$. Let $H$ be generated by $x^3$. Then $H$ has two elements, and $G/H$ has three elements. Since $G/H$ is cyclic of prime order, any nontrivial element generates it. Therefore, $x^2H$ is a generator of $G/H$.
But $x^2$ has order $3$ in $G$, so $x^2$ does not generate $G$.
This is the smallest counterexample possible (unless you allow $H=G$, which are you implicitly but not explicitly excluding): you cannot have a counterexample if $G$ has prime order (since in that case $H$ must be either trivial or the whole thing). And for the cyclic group of order $4$, all elements that do not lie in the nontrivial proper subgroup generate the whole group. So the smallest order where a counterexample was possible was $6$, and there it is.