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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $1$ (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
  • 1
    $\begingroup$ This paper may be of interest to you. $\endgroup$ – user170039 Aug 23 '16 at 13:16

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).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.