Application of Sylow Theorems to groups of order $pq^2r$ 
Suppose G is a group of order $pq^2r$, where $p,q,r$ are primes and $p  < q < r.$ Suppose further that $q^2 \not\equiv 1\pmod{r}$, $pq^2 \not\equiv 1\pmod{r}$, and $\gcd(q,r-1) = 1$. Show that either $G$ contains a unique Sylow $r$-subgroup, or that $Z(G)$ contains an element of order $q$.

Here is my attempt at the problem. Striving for a contradiction, suppose that neither $G$ has a unique Sylow $r$-subgroup, nor does $Z(G)$ have an element of order $q$. By the third part of Sylow's Theorem, $n_r \equiv 1\pmod{r}$ and $n_r \in \{1,p,q,q^2,pq,pq^2\}$. By our assumption that $G$ does not have a unique Sylow $r$-subgroup, we cannot have $n_r = 1$. We cannot have $n_r = p$, since $p \not\equiv 1\pmod{r}$, and similarly $n_r \neq q.$ Furthermore, by our hypotheses that $q^2, pq^2 \not\equiv 1\pmod{r}$, we deduce by exhaustion that $n_r = pq.$
Next, we look at possible values for $|Z(G)|.$ By Lagrange's Theorem, we have $|Z(G)| \in \{1,p,q,q^2,r,pq,pq^2,pr,qr,q^2r,pqr,pq^2r\}.$ By Cauchy's Theorem, combined with our assumption that $Z(G)$ contains no element of order q, we immediately reduce this set of possibilities to $\{1,p,r,pr\}.$ Now suppose $R \in \text{Syl}_r(G)$. Then $[G:N_G(R)] = n_r = pq$ implies $|N_G(R)| = qr$. Since $Z(G) \leq N_G(R)$, we must have $|Z(G)|$ divides $qr$. Thus $|Z(G)| \in \{1,r\}$. But if $|Z(G)| = r$, then $Z(G)$ is a normal Sylow $r$-subgroup, which would imply $n_r = 1$, a contradiction. We deduce that $G$ has trivial center.
Finally, if we look at $n_q$, we see that $n_q \in \{1,p,r,pr\}$. Since $n_q \equiv 1\pmod{q}$, we cannot have $n_q = p$. If $n_q = r$, then we may write $r = 1 + kq$ for some $k \in \mathbb{Z}$. But then $r-1 = kq$, which implies that $q = \gcd(q,r-q)$, a contradiction. Thus either $n_q = 1$ or $n_q = pr.$
From here I'm kind of stuck.
 A: if $n_{q}=1$ then let $Q<G$ be a sylow q-subgroup of $G$ and consider $\phi :G \to Aut(Q)$ s.t. for all $g \in G$ and $x \in Q$ we have $\phi (g)(x) = gxg^{-1}$, since $Q$ is normal we get that for all $g\in G$ the function $\phi (g)$ is really an automorphism on $Q$ From the first isomorphism theroem we deduce that $|G| = pq^2r = |ker(\phi)||Im(\phi)|$, also $Q$ is a $q$ group of order $q^2$ so $Q\cong \Bbb Z_{q^2}$ or $Q \cong \Bbb Z_{q} \times \Bbb Z_{q}$, and so $|Aut(Q)| = q(q-1)$ or $|Aut(Q)| = q(q-1)^2(q+1)$. From this follows that necceceraly $|Im(\phi)| = q,|Ker(\phi)| = pqr$  or  $|Im(\phi)| = 1,|Ker(\phi)| = pq^2r$. Also notice that $Q$ is abelian (group of order $q^2$) so $Q<Ker(\phi)$ and so $|Im(\phi)| = 1,|Ker(\phi)| = pq^2r$ and from here $Q<Z(G)$ (since every element in $G$ commutes with every element in $Q$)and clearly $Z(G)$ has an element with order $q$ (caushy).
A: You are off to a good start. I will jump in at the point were you had concluded that
if $n_r>1$ then $n_r=pq$ and $|N_G(R)|=qr$ for any Sylow $r$-subgroup $R$.
Let us fix one $R$. We know that there exists an element $z\in N_G(R)$ such that $\operatorname{ord}_G(z)=q$. The automorphism group of $R$ is cyclic of order $r-1$, so it has no elements of order $q$. Therefore we can conclude that $z$ actually centralizes $R$.
Getting warmer.
Next, we also know that $z$ is an element of at least one Sylow $q$-subgroup. Let $Q$ be one such. Because $Q$ has order $q^2$ we know that it is abelian. Also, the $Q$-orbit of conjugates of $R$ must have exactly $q$ Sylow $r$-subgroups. This is because

*

*$Q$ is not contained in $N_G(R)$, so the orbit has size $>1$, and

*$\operatorname{Stab}_R(Q)$ contains $z$, so the orbit has size $<q^2$.

Let $x\in Q$ be arbitrary. We have $C_G(xRx^{-1})=xC_G(R)x^{-1}$. It follows that
$xzx^{-1}\in C_G(xRx^{-1})$. But $Q$ is abelian, so we have shown that
$$
z\in C_G(xRx^{-1})\ \text{for all $x\in Q$.}
$$
Let $Z=C_G(z)$:

*

*Because $Q$ is abelian $q^2\mid |Z|$.

*Because $z$ centralizes $R$ we have $r\mid |Z|$, so $q^2r\mid |Z|$.

*But, we also know that $Z$ contains at least those $q$ conjugates of $R$. As $q^2\not\equiv1\pmod r$ the number of Sylow $r$-subgroups of $Z$ cannot be $1,q$ or $q^2$ either.

It follows that $Z$ must have more than $q^2r$ elements Hence $Z=G$, and $z\in Z(G)$.
