2
$\begingroup$

Let $G$ be a group which has a normal Sylow subgroup. Must every nontrivial quotient group $G/N$ have a nontrivial normal Sylow subgroup?

For example, every group of order up to 23 is a semidirect product of its Sylow subgroups so they all have a nontrivial normal Sylow subgroup, and also their quotients. $S_4$ doesn't have a normal Sylow subgroup. Is there a group which has a nontrivial normal Sylow subgroup but not every quotient has it?

Just finished posting the question and I noted it. Take $G=S_4\times \mathbb{Z}/7\mathbb{Z}$.

$\endgroup$
  • $\begingroup$ Wait, 23 is prime. $\endgroup$ – Randall Sep 8 '17 at 18:18
  • $\begingroup$ Yes. And $\mathbb{Z}/23\mathbb{Z}\cong \mathbb{Z}/23\mathbb{Z}\rtimes 1$. But anyway, I just found out what I was looking for (By the way, by nontrivial I mean different from $1$). $\endgroup$ – David Molano Sep 8 '17 at 18:19
0
$\begingroup$

You could take $\langle (12)(34), (13)(24)\rangle=K \lhd A_4.$ Since $|A_4|=12,$ and $|K|=4,$ then $K$ is a sylow $2-$subgroup. The quotient has no nontrivial normal Sylow subgroup since it's $\Bbb{Z}/3\Bbb{Z},$ and cyclic groups of prime order are generated by all of their non-identity elements.

$\endgroup$
  • $\begingroup$ Well, by $P$ being trivial I meant $P=1$, and not $P\in \{1,G\}$. As in the definition of Sylow Tower (Where we accept all abelian groups, in particular, $\mathbb{Z}/3\mathbb{Z}$ to have one). $\endgroup$ – David Molano Nov 7 '17 at 4:07
-1
$\begingroup$

Take any group $H$ which doesn't have a normal Sylow subgroup, take $p\nmid|H|$ and let $G=H\times C$ where $C$ is cyclic of order $p$. Then $G$ has a normal Sylow $p-$subgroup, but $G/(1\times C))\cong H$ doesn't. The example can be generalized even more, taking $C$ as any $p-$group.

$\endgroup$

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.