# 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 \not\equiv 1$$ (mod $$q$$), $$qr. 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.