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I'm doing 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 (in particular, $G/C$ is an abelian group).

Prove that $C=\langle[a,b]=aba^{-1}b^{-1} \mid a,b\in G \rangle=\{e\}$ and $G$ is abelian.

I really don't know how to prove this. By Lagrange I saw that the order of $C$ could be $1$ or $q$, but the option $|C|=q$ is still a valid option, so I don't know how to show that $C=\{e\}$. Thank you.

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  • $\begingroup$ Do you know Sylow's theorem(s)? If yes, prove that there is a unique (hence normal) $p$-Sylow subgroup $P$. First show that $P$ is central, then look at $G/P$. $\endgroup$ – j.p. Apr 2 '15 at 22:11
  • $\begingroup$ @j.p. If I see that $P$ is central ($P\leq Z(G)$) why can I say that that $G/Z(G)$ is cyclic? Also could you say me how to see that $P$ is central? $\endgroup$ – Relure Apr 3 '15 at 3:48
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    $\begingroup$ Given two nontrivial elements $x$ and $y$ of a group $G$, assume that $x$ has prime order $p$. Then there are two possibilities: Either $x$ and $y$ commute, i.e., $xy=yx$, or $y^x := x^{-1}yx \ne y$ (some define $y^x = xyx^{-1}$ instead) is a conjugate of $y$ different from $y$, and so are $y^{x^2}, y^{x^3}, \dots y^{x^{p-1}}$ all different elements. As $x$ has order $p$, $x^p=1$ and so $y^{x^p}=y$. So we have $p$ elements which are all conjugates to each other via $X:=\langle x\rangle$. The technical term is that $X$ acts via conjugation on $G$. Each orbit has length either $p$ or $\endgroup$ – j.p. Apr 3 '15 at 20:06
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    $\begingroup$ $1$ (which is the case if both elements commute). Now if $y$ is an element of $P$, all conjugates are in $P$ as $P$ is normal. OK, maybe I should not have called the prime $p$, but $r$ instead. Then you see, as $p\ne 1\pmod r$, that $P$ contains another element commuting with $x$ besides the obvious element $1$. But as $P$ is cyclic of prime order, this element generates $P$ and therefor all elements must commute with $x$. That's how you use these strange conditions that "everything" is not equal $1$ modulo "something else". $\endgroup$ – j.p. Apr 3 '15 at 20:12
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    $\begingroup$ This paper may be of interest to you. $\endgroup$ – user 170039 Aug 23 '16 at 13:16
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I found a document that solved this problem. It's a particular case, but it helps on understanding:

http://faculty.etsu.edu/gardnerr/4127/notes/VII-37.pdf (Pages 9 and 10, Example 37.15).

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