# Proving that a group of order $pqr$ (with conditions on those primes) is abelian.

I'm doing this exercise:

Let $$G$$ be a group, with $$|G|=pqr$$, $$p,q,r$$ different primes, $$q, $$r \not\equiv 1$$ (mod $$q$$), $$qr. 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.

• 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 $G/P$. – j.p. Apr 2 '15 at 22:11
• @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? – Relure Apr 3 '15 at 3:48
• 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 – j.p. Apr 3 '15 at 20:06
• $1$ (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". – j.p. Apr 3 '15 at 20:12
• This paper may be of interest to you. – user170039 Aug 23 '16 at 13:16