# Let $G$ be a group. Let $G/K$ be an abelian group. Prove that $C=\{e\}$.

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.

• You must be missing something, since $K = C$ satisfies your problem statement. – David Wheeler Apr 1 '15 at 20:32
• Yes, that's what I came up with too. I'll complete the question with some more information. Thanks. – Relure Apr 1 '15 at 20:36
• @DavidWheeler It changes something if $K$ is the unique Sylow $p$-subgroup of $G$? – Relure Apr 1 '15 at 20:43
• $G= D_6 \times C_5$, $p,q,r=3,2,5$ gives $C=K$. Also how is $H$ involved? – Derek Holt Apr 1 '15 at 20:58
• @DerekHolt There're more restrictions. I'm going to edit it and put the whole exercise because it's leading to confusions. – Relure Apr 1 '15 at 21:01

$\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).$$• Hi @Andreas . There's a way to prove it without using the centralizer? I think that they want us to deduce it, like if it was easier than this. – Relure Apr 1 '15 at 21:23 • @Abrahamlure, perhaps you have further information about$H$, which at the moment, as noted by Derek Holt, plays no role? – Andreas Caranti Apr 1 '15 at 21:27 • ... I'm so sorry. I use to name$H$to a subgroup of$G$. Every$H$is a$K$. It's edited now. – Relure Apr 1 '15 at 21:29 • I think that the important thing here is that$K$is the unique Sylow$q$-subgroup. There's some way to say that$C\neq K$because$K$is the unique Sylow$q\$-subgroup? – Relure Apr 1 '15 at 21:52