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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.