Group of order $|G|=pqr$, $p,q,r$ primes has a normal subgroup of order

Question

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.

Answer 1

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.

Answer 2

More simply, since any square-free group is solvable, in any group of square-free order the largest Sylow subgroup (or order $r$) is normal by Philip Hall's theorem (since the number of them must be a product of numbers congruent to $1\pmod{r}$ and each dividing a chief factor, but all the chief factors are primes $\le r$). So we have $P_r\trianglelefteq G$, and $G/P_r$ by the same reasoning has a normal subgroup of order $q$, and its preimage in $G$ is the desired normal subgroup.

Similarly, for any group of order $p_1p_2\cdot\cdot\cdot p_n$ with $p_1<p_2<\cdot\cdot\cdot <p_n$ primes, there are normal subgroups of orders $p_n$, $p_np_{n-1}$, ... $p_np_{n-1}\cdot\cdot\cdot p_2$.