A group of order $pqr$ (primes $p > q > r$) has a subgroup of order $qr$ I've done most of the following problem, but I can't seem to get part (d).

Let $G$ be a group of order $pqr$ for primes $p > q > r$.  By a counting argument one can see that there is either a normal Sylow $p$-subgroup or a normal Sylow $q$-subgroup (you may assume this).  Prove:
(a) $G$ has a normal subgroup $H$ of order $pq$.
(b) Every subgroup of order $p$ or $q$ is contained in $H$.
(c) $G$ has a normal subgroup of order $p$.
(d) $G$ has a subgroup of order $qr$.

Proof: Let me first state the following lemma: if $N_0$ is a subgroup of a finite group $G_0$, and $[G_0 : N_0]$ is equal to the smallest prime which divides $|G_0|$, then $N_0$ is normal in $G_0$.  The proof is not difficult.  It can be established by a counting argument, or by looking at the natural homomorphism from $G_0$ to the symmetric group on the left cosets of $N_0$ in $G_0$.
For (a), let $N$ be a Sylow $p$-subgroup, and $K$ a Sylow $q$-subgroup.  One of these subgroups is normal by assumption.  So $H := NK$ is a subgroup of $G$ with order $pq$.  Its index in $G$ is $r$, so it is normal in $G$ by the lemma.
(b) If $P$ is a subgroup of order $p$, it is a Sylow $p$-subgroup, so it equals $gNg^{-1}$ for some $g \in G$.  So $g^{-1}Pg = N \subseteq H$, whence $P \subseteq gHg^{-1} = H$. Similarly for a subgroup of order $q$.
(c) Let $P$ be a subgroup of order $p$.  Then it is a Sylow $p$-subgroup and it is contained in $H$. Since $|H| = pq$, and $[H : P] = q$, automatically $P$ is normal in $H$.  If $P_1$ is another Sylow $p$-subgroup of $G$, then it is also contained in and normal in $H$, whence $P = P_1$.
(d) Let $Q$ be a Sylow $q$-subgroup, and $R$ a Sylow $r$-subgroup.  If $Q$ is normal in $G$, we are done, since $QR$ is a subgroup of order $qr$.  Otherwise, $n_q$ (the number of Sylow $q$ subgroups) is either $r, p,$ or $pr$.  We cannot have $n_q = r$, since then $q$ divides $r-1 < q$.  If $n_q = p$, then $|N_G(Q)| = qr$ and we are done.  The last possibility is that $n_q = pr$.  I don't know what to do here.  I've tried making a counting argument involving the possibilities for $n_r$, but I haven't been able to make anything work.
 A: D follows easily from the following two theorems:


*

*Every group of square-free order is solvable.

*Hall Theorem: A solvable group admit all the possible Hall subgroups.

A: Ofir's answer is a high powered solution, and here is a more elementary approach as suggested by Derek.
Proof of (d): Let $Q$ be a Sylow $q$-subgroup of $G$.  We first claim (this is the "Frattini argument")that $$G = N_G(Q)H$$ If $g \in G$, we know that $Q \subseteq H$ (therefore a Sylow $q$-subgroup of $H$), so $gQg^{-1} \subseteq gHg^{-1} = H$, since $H$ is normal in $G$.  Since $gQg^{-1}$ is a Sylow $q$-subgroup of $H$, it is equal to $hQh^{-1}$ for some $h \in H$.  Then $h^{-1}gQg^{-1}h$, or $h^{-1}g \in N_G(Q)$.  So $g \in HN_G(Q) = N_G(Q)H$.  
Therefore, $$|G| = |N_G(Q)H| = \frac{|N_G(Q)| \cdot |H|}{|N_G(Q) \cap H|} = \frac{|N_G(Q)| \cdot pq}{|N_G(Q) \cap H|}$$ so $n_q = [G :N_G(Q)] = \frac{pq}{|N_G(Q) \cap H|}$.  The denominator is divisible by $q$, so $n_q$ is either $1$ or $p$.  
If $n_q = 1$, then $Q$ is normal in $G$ and $QR$ is a subgroup of order $qr$, where $R$ is a Sylow $r$-subgroup.  If $n_q = p$, then $N_G(Q)$ has order $qr$ and is the desired subgroup.  
