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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?

citadel
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abcdmath
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    Do you have the assumption that $G$ is abelian? – Kal S. Nov 04 '17 at 22:05
  • @Test123. No there is no assumption that $G$ is abelian. Then I know that $G$ must be cyclic. – abcdmath Nov 04 '17 at 22:07
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    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 $r-1$. – Derek Holt Nov 04 '17 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 $r-1$. – abcdmath Nov 04 '17 at 22:37
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    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 $pqr$. – Derek Holt Nov 05 '17 at 08:20
  • A link to two special cases: order 255 https://math.stackexchange.com/q/255441/61691 ; order 455 https://math.stackexchange.com/q/2988965/61691 – azimut Aug 23 '22 at 15:38

2 Answers2

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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 $pq$.

Kal S.
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  • This is nice and totally new to me. Thanks for the help. – abcdmath Nov 04 '17 at 22:33
  • Can you provide me the proof of the proposition in your answer? – abcdmath Nov 04 '17 at 22:43
  • Is there a proof of this proposition which does not use Sylow theorems? – Raito Nov 04 '17 at 22:44
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    @abcdmath Here is a proof, which uses Sylow theorem and results on Euler's totient function. https://yiminge.wordpress.com/2009/01/22/all-groups-of-order-n-are-cyclic-iff/ – Raito Nov 04 '17 at 22:44
  • Thank you sir for your help. – abcdmath Nov 04 '17 at 22:47
  • @abcdmath The proof of the proposition uses Sylow theorems. In general proving that a finite group of order $n$ is cyclic for specific values of $n$, requires the use of Sylow theorems. – Kal S. Nov 04 '17 at 23:09
  • The proposition is different: Every group of order $n$ is cyclic iff this holds. Otherwise there are counterexamples. Take $n=pq$ with the distinction $p\mid q-1$ or $p\nmid q-1$. – Dietrich Burde Nov 05 '17 at 10:08
  • @abcdmath More concretely, take $n=21$ and the abelian group of order $21$. We have $gcd(21,\phi(21))=3$, so the cyclic group is not cyclic according to this Proposition! – Dietrich Burde Nov 05 '17 at 10:13
  • @DietrichBurde I didn't fully understand what you mean. For now I edited the proposition to the one direction only. I can't recall the full statement though. What is the correct statement? – Kal S. Nov 05 '17 at 11:26
  • @Test123 Click on Raito's link above for the correct statement in both directions. Yours is fine, too, for this purpose, of course. – Dietrich Burde Nov 05 '17 at 12:21
  • @DietrichBurde Oh now I understand what you meant earlier. Thanks. I haven't realised that my initial statement missed the subtlety of 'every' group. I thought that the one direction was not true after reading your earlier comment. – Kal S. Nov 05 '17 at 13:38
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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 $qr$ $p$-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 $pqr-r(p+q)+1>$ $pqr-2qr+1=$ $qr(p-2)+1>0$.

citadel
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