I'm stuck at this exercise:
Let $G$ be a group, with $|G|=pqr$, $p,q,r$ different primes, $q<r$, $r \not\equiv 1$ (mod $q$), $qr<p$. Also suppose that $p \not\equiv 1$ (mod $r$), $p \not\equiv 1$ (mod $q$). Let $C$ (the commutator of $G$) and $K$ be subgroups of $G$, with $C \leq K$, $K \trianglelefteq G$ and $|K|=q$. $K$ is the unique Sylow $q$-subgroup on $G$ (so $K \trianglelefteq G$). Let $G/K$ be an abelian group. Prove that $C=\{e\}$.
I tried using Lagrange theorem, knowing that $C\leq K$, and then $|C|=\{1\ or\ q\}$. But I don't know how to eliminate the option $|C|=q$.
This is a little part from a longer exercise. The definition of $C$ is $C=\langle[a,b]=aba^{-1}b^{-1} \mid a,b\in G \rangle$, $G/C$ is abelian too.
Thank you.
$\newcommand{\Span}[1]{\langle#1\rangle}$$G/K$ is abelian of order $p r$, thus cyclic. If you can prove that $K \le Z(G)$, then it follows that $G$ is abelian.
To prove this, let $K = \Span{z}$. We have that $G/C_{G}(K)$ has order dividing $q-1$.
Let $a$ be an element of order $p$, and $b$ an element of order $r$. Since $p > q r > q$, we have that $a$ centralizes $z$. Since $r > q$, we have that $b$ centralizes $z$. Hence $z$ is central, as required.
Lemma. Let $G$ be a group. Suppose $G/Z(G)$ is cyclic. Then $G$ is abelian.
Proof. Suppose $G = \langle a, Z(G) \rangle$. Then two arbitrary elements of $G$ can be written as $a^{i} z, a^{j} w$, with $i, j$ integers, and $z, w \in Z(G)$. Thus $$ (a^{i} z) (a^{j} w) = a^{i} a^{j} z w = a^{j} a^{i} w z = (a^{j} w) (a^{i} z). $$