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I'm stuck on this exercise:

Let $G$ be a group, with $|G|=pqr$, $p,q,r$ different primes, $q<r$, $r \not\equiv 1$ (mod $q$), $qr<p$. Show $G$ has an unique Sylow $p$-subgroup $P$.

What I've done:

We know, by the first Sylow theorem, that $P$ exists, and $|P|=p$. By the third Sylow theorem, we have that $n_p$ (the number of Sylow $p$-subgroups) has to verify that: $$n_p \mid qr$$

and $$n_p\equiv 1 \text{(mod p)}.$$

So, $n_p=\{1,q,r,qr\}$ using that $\ n_p \mid qr$.

First, we see that $n_p\neq qr$, because $qr<p$, and then $qr \not\equiv1$ (mod $p$).

I would appreciate any hint to eliminate the options $n_p=q$ and $n_p=r$.

Thank you.

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    $\begingroup$ If $qr < p$, then also $q<p$ and $r<p$, so neither $q$ nor $r$ can be congruent to $1$ modulo $p$. $\endgroup$ – James Mar 30 '15 at 18:42
  • $\begingroup$ You're right. Thank you @James, I don't know how I overlooked that. Put it as an answer and I'll accept it. $\endgroup$ – Relure Mar 30 '15 at 18:47
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Since $qr < p$, then $q < p$ and $r < p$, so none of $q$, $r$ and $qr$ can be congruent to $1$ modulo $p$. Therefore, $n_p = 1$, as desired.

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