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.
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.