Let $p,q,r$ be positive primes, $p<q<r$, and let $G$ be a group with $|G|=pqr$. Show that there exists a normal subgroup $H$ of $G$ of order $qr$.
I've seen this post Groups of order $pqr$ and their normal subgroups with the same problem, but I couldn't follow the answer. I'll write what I've done up to now:
We know that $$n_r \equiv 1 (r),\ n_r\mid pq,\ r>q,p \implies n_r \in \{1,pq\}.$$ By a similar argument, we arrive to $$n_q \in \{1,r,pr\},$$ $$n_p \in \{1,q,r,qr\}.$$ Here $n_i$ denotes the number of $i$-Sylow groups for $i=p,q,r$.
If $n_r=1=n_q$, then there are only one $r$-Sylow subgroup, $H$ and one $q$-Sylow subgroup, $K$, then $H,K \lhd G$, so $KH$ is a normal subgroup and $|KH|=qr$.
I don't know what to do for other cases, I would appreciate some help.
Best Answer
Let $p,q,r$ be positive primes, $p<q<r$, and let $G$ be a group with $|G|=pqr$. We know that $$n_r \equiv 1 \bmod r,\ n_r\mid pq,\ r>q,p \implies n_r \in \{1,pq\}.$$ For the worst case take $n_{r}=pq$.
By a similar argument, taking the worst case we arrive to $$n_q =r$$ $$n_p =q.$$ Here $n_i$ denotes the number of $i$-Sylow groups for $i=p,q,r$.
Now by counting we see there are $pqr-pq+1$ elements of order $r$ (with $e$), and $qr-r$ elements of order $q$. Then show that either $q$-Sylow group or $r$-Sylow group is normal. Then construct $HK$. Then the result follows.