# 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?

• Do you have the assumption that $G$ is abelian? Commented Nov 4, 2017 at 22:05
• @Test123. No there is no assumption that $G$ is abelian. Then I know that $G$ must be cyclic. Commented Nov 4, 2017 at 22:07
• A group of order $pqr$ with $p,q,r$ primes and $p<q<r$ is necessarily cyclic if and only if $p$ does not divide $q-1$ or $r-1$, and $q$ does not divide $r-1$. Commented Nov 4, 2017 at 22:23
• @Derek Holt. Sir you have commented that "if $p$ does not divide $q-1$ or $r-1$" . Is it 'or'? Not 'and'? I mean to say that if $p$ does not divide both of $q-1$ and $r-1$. Commented Nov 4, 2017 at 22:37
• Sorry for any ambiguities. The group is necessarily cyclic if and only if all of the following three conditions hold: (i) $p$ does not divide $q-1$; (ii) $p$ does not divide $r-1$; (iii) $q$ does not divide $r-1$. In other words, if any one of those three conditions fails, then there is a non-cyclic group of order $pqr$. Commented Nov 5, 2017 at 8:20

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 $pq$.

• This is nice and totally new to me. Thanks for the help. Commented Nov 4, 2017 at 22:33
• Can you provide me the proof of the proposition in your answer? Commented Nov 4, 2017 at 22:43
• Is there a proof of this proposition which does not use Sylow theorems? Commented Nov 4, 2017 at 22:44
• @abcdmath Here is a proof, which uses Sylow theorem and results on Euler's totient function. yiminge.wordpress.com/2009/01/22/… Commented Nov 4, 2017 at 22:44
• Thank you sir for your help. Commented Nov 4, 2017 at 22:47

WLOG, let's assume $$p.

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 $$qr$$ $$p$$-Sylow subgroups and $$pr$$ $$q$$-Sylow subgroups. Since all these subgroups intesect pairwise trivially (they have order $$p$$ or $$q$$), their union's size amounts to $$qr(p-1)+pr(q-1)+1$$, which is greater than $$pqr^\dagger$$. 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 $$p\nmid r-1$$.

$$^\dagger$$In fact: $$qr(p-1)+pr(q-1)+1=$$ $$2pqr-r(p+q)+1>pqr\iff$$ $$pqr-r(p+q)+1>0$$, which is true because $$pqr-r(p+q)+1>$$ $$pqr-2qr+1=$$ $$qr(p-2)+1>0$$.