Any group of order $n$ satisfying $\gcd (n, \varphi(n)) =1$ is cyclic Sorry for the last mistaken problem I just posted. Now I know that only having the order being odd square free is not enough for a group to be cyclic. 
Here's the complete problem which the main goal is to show that any group of order $n$ satisfies $\gcd (n,\varphi(n))$=1 is cyclic. It asks me to do it in the following way:
(a) If $n > 2$, $n$ is an odd squarefree integer.
(b) Show that there is a $H = G/N$ of prime order.
(c) Show that $G \cong N \times H$ and $G$ is abelian.
(d) $G$ is cyclic.

Any ideas is appreciated. 
 A: The proofs of a), b), d) seem correct. To prove c) notice first that if you take an element $x\in G$ such that order of $x$ equals $p_{i_n}$, you have $\{e/N,x/N,\ldots,x^{p_{i_n}-1}/N\}=G/N$, so the exact sequence $0\to N\to G\to H\to 0$ splits and $G\cong N\rtimes H$.
Therefore, it's enough to show that $H$ can't act nontrivially on $N$. But by induction (with respect to number of prime factors of $G$) you can assume that $N\cong {\bf Z}_{\lvert N\rvert}$ which has automorphism group isomorphic to ${\bf Z}_{\lvert N\rvert}^*$.  
From that, an action of $H$ on $N$ induces a homomorphism from $H$ to ${\bf Z}_{\lvert N\rvert}^*$. But orders of $H$ and ${\bf Z}_{\lvert N\rvert}^*$ are coprime by assumption, so the only action is the trivial one and we're done.
Note that the converse of the theorem you are to prove is also true: if $n, \varphi(n)$ are not coprime, then there's a group of order $n$ which is not cyclic: if $n$ is not squarefree, then this is trivial (${\bf Z}_{n/p}\times {\bf Z}_{p}$ has order $n$ and is non-cyclic). Otherwise, if you take $p$ such that $p$ divides both $n$ and $\varphi(n)$, then $p$ divides $\varphi(n/p)$ and you have a nontrivial semidirect product ${\bf Z}_{n/p}\rtimes {\bf Z}_p$.
