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.

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

  • $\begingroup$ 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. $\endgroup$ – Relure Apr 1 '15 at 21:23
  • $\begingroup$ @Abrahamlure, perhaps you have further information about $H$, which at the moment, as noted by Derek Holt, plays no role? $\endgroup$ – Andreas Caranti Apr 1 '15 at 21:27
  • $\begingroup$ ... I'm so sorry. I use to name $H$ to a subgroup of $G$. Every $H$ is a $K$. It's edited now. $\endgroup$ – Relure Apr 1 '15 at 21:29
  • $\begingroup$ 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? $\endgroup$ – Relure Apr 1 '15 at 21:52
  • 1
    $\begingroup$ @Abrahamlure, no way that way. $\endgroup$ – Andreas Caranti Apr 1 '15 at 21:54

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