Let $G$ be group of order $pqr$, when $p,q,r$ are different prime numbers. Does $G$ must have normal cyclic subgroup $H$ such that $G/H$ is cyclic too ?

I know that $G$ has normal sylow subgroup of order $p$ or $q$ or $r$, and that $G$ is solvable, but I can't see why there is a cyclic quotient.

Edit: Attempt for solution - If $G$ is abelian, then $G\cong \mathbb{Z}_{p}\oplus\mathbb{Z}_{q}\oplus\mathbb{Z}_{r}\cong \mathbb{Z}_{pqr}$, hence cyclic and it's clear. If not, then $G'\neq \left\{ e\right\}$. If $\left|G'\right|=p$, then $G/G'\cong \mathbb{Z}_{qr}$ (because $G'/G$ is abelian), both cyclic. and the other cases are symetric. $G'\neq G$ because $G$ is solvable.


1 Answer 1


The answer is yes. Let's suppose that $G$ has a normal Sylow $p$-subgroup $P$.

If $C_G(P) = P$ then, since $G/C_G(P)$ is isomorphic to a subgroup of ${\rm Aut}(P)$, which is cyclic, $G/P$ is cyclic.

If $P < C_G(P) < G$, then $C_G(P)$ and $G/P$ are both cyclic.

Finally, if $C_G(P) = G$ then $G/P$ has a normal Sylow subgroup $K/P$ say, and then $K$ and $G/K$ are cyclic.

  • $\begingroup$ why does $C_{G}\left(P\right)$ cyclic in the second case ? I added my attempt for solution, I would be glad if you check it. $\endgroup$
    – daPollak
    Jan 24, 2015 at 21:01
  • $\begingroup$ In the second case $C_G(P)/P$ is cyclic so $C_G(P)$ is abelian and hence cyclic. $\endgroup$
    – Derek Holt
    Jan 24, 2015 at 22:39
  • $\begingroup$ In your solution, you don't appear to have considered the case when $|G'|=pq$. $\endgroup$
    – Derek Holt
    Jan 24, 2015 at 22:43
  • $\begingroup$ Yup you're right. Do you have an idea for a solution using the fact that $G$ is solvable ? $\endgroup$
    – daPollak
    Jan 25, 2015 at 0:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .