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Assume $|G| = pqr$ where $p,q,r$ are primes with $p < q < r$. Then $G$ is not simple.

I have a problem understanding the proof (see for example here). In the proof one assumes that $n_p,n_q,n_r > 1$ (number of each $p,q,r$-Sylow subgroups respectively) and then by Sylow we have $$n_r | pq \qquad \text{and} \qquad n_r = 1 + kr, k\in \mathbb{N}_0$$ Now one deduces that $n_r = pq$, which I do not understand.

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TheGeekGreek
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1 Answers1

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Since $n_r$ is not $1$, we have $k>0$ which implies $n_r=1+kr > r > p$ and $q$, so the only possible divisor of $pq$ is $pq$.

Tara
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