I'm trying to prove the following result. I think I'm not understanding it correctly. Just to clarify, my knowledge extends to Sylow's theorem corollaries but I don't know about solvable groups.

Let $G$ be a group of order $pqr$, $p>q>r$. Prove that $|G| \geq 1 + n_p(p-1) + n_q(q-1) + n_r(r-1)$

Here's what I have thought.

What I understood is that equation tells me to prove that there are more elements of the group $G$ than elements that belong to any Sylow subgroup. Isn't this always valid? Which means that I'm looking at the case when the inequality is a strict one?. This confuses me. Anyways, in that case if I find an element of order, say $pq$, then that element will not be an element of any Sylow subgroup. So the equation will be valid with a strict inequality.

I know that in this group there is a normal group of order $p$ which means that $n_p = 1$. From this I can conclude that there is a normal subgroup of order $pq$. If I knew that $p \neq 1\mod{q}$ , then the subgroup would be a cyclic subgroup which means that I can find an element of order $pq$ and the equation will be valid. Bu there are no restrictions on either $p$ or $q$.

I would appreciate any hints. Sorry if it has been asked before.

Edit: Maybe the question could be rephrased as is there a cyclic subgroup of order bigger than $p$? If that's true then the problem is solved.

  • $\begingroup$ You are right, in this case this inequality just says that $G$ has more elements than in Sylow's subgroups, and this is always true. $\endgroup$ – SMM Jul 8 '18 at 15:31

Since all Sylow subgroups of $G$ are of prime order, the intersection of any two distinct Sylow subgroups is trivial. Hence there are $n_p (p − 1)$ elements of order $p$, $n_q (q − 1)$ elements of order $q$, $n_r(r − 1)$ elements of order $r$, and $1$ element of order $1$ in G. Therefore the inequality follows.

  • $\begingroup$ I see. Thanks for the answer! Do you know when the inequality is strict? I have tried some small cases and found that the inequality is strict but couldn't quite prove it in the general case. $\endgroup$ – Leo Lerena Jul 8 '18 at 17:33
  • 2
    $\begingroup$ The inequality is strict, because all such groups have elements of composite order. The proof depends on the non-obvious (and not that easy) theorem that if all Sylow subgroups are cyclic, then $G$ is a semidirect product of two cyclic groups. Obviously one of those factors must be a cyclic subgroup of composite order, so it has elements of composite order. $\endgroup$ – C Monsour Jul 8 '18 at 18:25
  • $\begingroup$ @CMonsour Thanks a lot for the answer. Just I was thinking how to prove that theorem in the special case that the group order is $231=3*7*11$, i.e that it's isomorphic to a semi direct product of $\mathbb{Z}_{33}$ and $\mathbb{Z}_7$. $\endgroup$ – Leo Lerena Jul 8 '18 at 18:46
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    $\begingroup$ Well, the 7 and 11 Sylows need to be normal, and the cyclic group of order 11 has no automorphisms of order 3, so the only non-abelian possibility is $\Bbb{Z}/(7)\rtimes\Bbb{Z}/(33)\cong\Bbb{Z}/(77)\rtimes\Bbb{Z}/(3)\cong F_{21}\times \Bbb{Z}/(11)$, where $F_{21}$ is the non-abelian group of order 21. $\endgroup$ – C Monsour Jul 8 '18 at 18:54
  • $\begingroup$ @CMonsour Thanks a lot!! You make it look so easy. $\endgroup$ – Leo Lerena Jul 8 '18 at 19:52

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