[Math] 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?

Best Answer

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 $pq$.

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