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. $\gcd(k+ln,m)=1$?
Could you please help me with this problem? Any help is really appreciated.
Best Answer
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 $G/H$.