We have $N\unlhd G$, a cyclic subgroup of order $n$, $G/N$ cyclic group of order $m$ s.t. $\gcd(m,\phi(n))=1$. Show $G$ is abelian.

Question

Problem statement:

We have $N \trianglelefteq G$ a normal cyclic subgroup of order $n$, $G/N$ cyclic group of order $m$ such that $\gcd(m, \phi(n)) = 1$, where $\phi$ is the Euler function. Show that $G$ is abelian.

My thoughts:

1. Take the generators $aN$ of $G/N$ and $x$ of $N$. Now if we can show that $ax = xa$, then we are done.
2. $\phi(n)$ seems to be pointing to the size of set of generators in $N$. But I am stuck here...

Answer 1

The group $G$ acts by conjugation on $N$. This yields a homomorphism $f:G\to\mathrm{Aut}(N)\cong(\mathbb{Z}/n\mathbb{Z})^\times$. Since $N$ is abelian, $N$ is contained in the kernel of $f$. By hypothesis, $|G/N|=m$ and $|(\mathbb{Z}/n\mathbb{Z})^\times|=\phi(n)$ are coprime. Hence, $\ker f=G$, i.e. $N$ is contained in the center of $G$. Now you can use your first argument to show that $G$ is abelian.