Order of cyclic groups and the Euler phi function

According to Wikipedia, a cyclic number (in group theory) is one which is coprime to its Euler phi function and is the necessary and sufficient condition for any group of that order to be cyclic. Why is that true?

I can see that if $n$ is prime, that guarantees any group of order $n$ is cyclic, but I don’t seem to see how to extend it to $(n,\phi(n))=1$

It would be nice if someone could explain it to me. Thanks.

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Suppose $\gcd(n, \phi(n)) > 1$. If $n$ is not squarefree, there exists a prime $p$ such that $p^2$ divides $n$. Then $H = \mathbb{Z}_p \times \mathbb{Z}_p$ is not cyclic and neither is $H \times \mathbb{Z}_{\frac{n}{p^2}}$. If $n$ is squarefree, there exist prime divisors $p$ and $q$ of $n$ such that $q$ divides $p-1$. Then there exists a non-abelian group $H$ of order $pq$ and $H \times \mathbb{Z}_{\frac{n}{pq}}$ is not cyclic.

The other direction is not so easy to answer, but I’ll give you a few good references.

Jungnickel and Gallian give elementary proofs (or at least rough outlines for a proof) in these two papers:

Jungnickel, Dieter. On the Uniqueness of the Cyclic Group of Order $n$. Amer. Math. Monthly, Vol. 99, No. 6 (1992) JSTOR

Gallian, J. A. Moulton, David. When is $\mathbb{Z}_n$ the only group of order $n$?, Elemente der Mathematik, Vol. 48 (1993) Link to article

Pakianathan and Shankar have a paper that goes beyond cyclic numbers and gives a number theoretic characterization for abelian, nilpotent and solvable numbers too:

Pakianathan, Jonathan. Shankar, Krishnan. Nilpotent Numbers. Amer. Math. Monthly, Vol. 107, No. 7 (2000) JSTOR