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.

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

• Thank you, How might I read the jstor articles? I am happy to pay but it says "Unable to process payment. Please contact the merchant as the postal address provided by the merchant is invalid, and the merchant has requested that your order must be delivered to that address." :( – retro Feb 5 '12 at 22:24
• @Retro: If you are studying at a university, you can probably get free access from there. Gallian's article can be found for free online, I added a link. – Mikko Korhonen Feb 6 '12 at 6:47
• Thank you very much for the link! I still don't know why I can't use JSTOR but the link is good :) – retro Feb 6 '12 at 12:18