# The number of Sylow subgroups on $G$ with $|G|=pqr$

I'm doing a part of an exercise and I don't know how to go on. Here it goes:

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 $P$. Also suppose that $p \not\equiv 1$ (mod $r$), $p \not\equiv 1$ (mod $q$).

Prove that the number of subgroups of order $r$ in $G$ is $1$ or $pq$. Prove that the number of subgroups of order $q$ in $G$ is $1$ or $pr$.

Here's what I've done:

They're asking us to find the number of Sylow $r$-subgroups and the Sylow $q$-subgroups in $G$. By the third Sylow theorem, we have that $n_r$ (the number of Sylow $r$-subgroups) has to verify that: $$n_r \mid pq$$

and $$n_r\equiv 1 \text{(mod r)}.$$

By the assumption of $n_r \mid pq$, we have that $n_r$ can be $1,p,q$ or $pq$. It's easy to discard the options $p$ and $q$, so $n_r$ can be $1$ or $pq$.

Here is where I'm stuck. How can we prove that $n_r=pq$ is a valid option? How to prove that $$pq\equiv 1 \text{(mod r)}?$$

I also quit at this point proving that $n_q=1$ or $pr$. I would appreciate any hint. Thank you.

## 1 Answer

I'd say that you're done with the first proof. You know that $n_r$ is either $1$ or $pq$ as you've already ruled out $p$ and $q$. It's similar to saying that $2+2$ is either $4$ or $5$. You don't have to show that both options are valid, that's not the question. If you still want to try, see if you can find three primes $p$, $q$ and $r$ which satisfy all the conditions.