# Lift of a generator of a non-trivial quotient of a cyclic group with prime power order must be a generator of the whole group

Let $$G$$ be a cyclic group with prime power order and $$H$$ be a proper subgroup of $$G$$.

Problem: Any element in $$G$$ whose image in the cyclic quotient $$G/H$$ is a generator, is already a generator of $$G$$.

Approach: Let $$G = \langle \sigma \rangle$$ and $$|G| = p^m$$ be a prime power.

Because $$H$$ is a proper subgroup of G, there exists an $$n$$ which is not coprime to $$p$$ such that $$H = \langle \sigma^n \rangle$$. Then $$G/H = \{ H, \sigma H, \dots, \sigma^{n-1}H \}$$. The generators of $$G/H$$ are of the form $$\sigma^k H$$ where $$1 \leq k \leq n-1$$ and $$\gcd(k,p)=1$$.

How can I show that any lift $$\sigma^{k+ln}$$ of $$\sigma^k H$$ is a generator, i.e. $$\gcd(k+ln,m)=1$$?

We know that $$G$$ is cyclic and has order $$p^m$$, so we can without loss of generality assume that $$G = \mathbb{Z}/p^m\mathbb{Z}$$.
Now, recall that $$x$$ is a generator of some group $$\mathbb{Z}/n\mathbb{Z}$$ if and only if it generates $$1$$, the non trivial implication given by the fact that if $$cx = 1$$ then $$(mc)x = m(cx) = m$$ for any $$m \in \mathbb{Z}/n\mathbb{Z}$$. And $$x$$ generates $$1$$ if and only if there exists $$c$$ in the group such that $$cx = 1$$, that is, $$x$$ has an inverse modulo $$n$$. This condition is equivalent as saying that $$x$$ is coprime with $$n$$.
Thus, $$x$$ is a generator of $$G$$ if and only if it is coprime with $$p^m$$, and $$[x] \in G/H$$ will be a generator if and only if $$[x]$$ (or equivalently $$x$$) is coprime with $$p^s = |G/H|$$ for $$s < k$$. This is because subgroups of $$\mathbb{Z}/p^m\mathbb{Z}$$ are isomorphic to some $$\mathbb{Z}/p^r\mathbb{Z}$$ and so $$G/H \simeq \mathbb{Z}/p^{k-r}\mathbb{Z}$$.
Thus, by the contrapositive: if $$x$$ is not a generator of $$G$$, then it is not coprime with $$p^m$$, which is to say that $$x$$ is divided by $$p$$. Thus, $$x$$ and $$p^s$$ share $$p$$ as a common divisor, proving that $$[x]$$ is not coprime with $$p^s$$ and consequently, it is not a generator of $$G/H$$.