How to say that any group of order $pqr$ is cyclic with a provided relation between $p$, $q$, $r$, where they are all distinct primes. I have given that, let $G$ be a group of order $455$. Then I have to show that $G$ is cyclic. Then by using Sylow-theorems it can be solved. But my question is, is there any method to see quickly that a group of order $pqr$ with a relation between them, where all of $p$, $q$, $r$ are distinct primes, is cyclic. Just like a group of order $pq$ is cyclic if $q>p$ and $p$ does not divide $q-1$. For example, any group of order $15$ is cyclic just from the above argument. Here $455=5.7.13$. So is there any simple way to say $G$ is cyclic?
 A: 
Proposition: Let $n\in\mathbb{N}$. Then $\gcd(n,\varphi(n))=1,$ where $\varphi$ denotes the
  Euler’s totient function, if and only if every finite group of order $n$ is cyclic.

The condition $\gcd(n,\varphi(n))=1$, implies a unique finite group of order $n$ which then has to be cyclic. 
For $n=455$ we have that $$\varphi(455)=\varphi(5)\varphi(7)\varphi(13)=(5-1)(7-1)(13-1)=288$$
$$\gcd{(288,455)=1}$$
EDIT: Note that for $p<q<r$ distinct primes, $$\varphi(pqr)=(p-1)(q-1)(r-1)$$
So $\gcd(pqr,(p-1)(q-1)(r-1))=1,$ is equivalent to $p\nmid (q-1)$, $p,q\nmid (q-1)$, analogously with the case of groups of order $pq$.
A: WLOG, let's assume $p<q<r$.
By Sylow III, $n_p\mid qr$ and $n_p\equiv 1\pmod p$. So, $n_p=1,q,r,qr$ and $n_p=1+kp$. If $p\nmid q-1$ and $p\nmid r-1$, then $n_p\ne q,r$ and hence $n_p=1,qr$. Likewise, $n_q\mid pr$ and $n_q\equiv 1\pmod q$. So, $n_q=1,p,r,pr$ and $n_q=1+lq$. Now, $q\nmid p-1$ (because $q>p$); so, if $q\nmid r-1$, then $n_q\ne p,r$ and hence $n_q=1,pr$.
Suppose there are $qr$ $p$-Syolw subgroups and $pr$ $q$-Syolw subgroups. Since all these subgroups intesect pairwise trivially (they have order $p$ or $q$), their union's size amounts to $qr(p-1)+pr(q-1)+1=$ $2pqr-r(p+q)+1>pqr\iff$ $pqr-r(p+q)+1>0$, which is true because $pqr-r(p+q)+1>$ $pqr-2qr+1=$ $qr(p-2)+1>0$. Therefore, there isn't enough room in $G$ for so many $p$-Sylows and $q$-Sylows at the same time, and hence either $n_p=1$ or $n_q=1$.
If $n_p=1$, then the only $p$-Sylow is normal, say $H$. As such, $H$ is the union of conjugacy classes of $G$, with as many singletons as the order of $|H\cap Z(G)|$. But, by Lagrange, $|H\cap Z(G)|=1,p$, and the former option is ruled out because there aren't conjugacy classes (of sizes $p$, $q$, $r$ and their pairwise products) "filling the gap" $p-1$. Therefore $H$ is central, and being $G/H$ cyclic (as $q\nmid r-1$), then $G$ is abelian and finally cyclic (take the product of any three elements of order $p$, $q$ and $r$, respectively).
Same argument in case of $n_q=1$ (say $K$ the only $q$-Sylow, normal), with the only difference that now "filling non-centrally the gap" $q-1$ is prevented by the assumption $p\nmid q-1$, and the ciclicity of $G/K$ is ensured by $p\nmid r-1$.
