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 $G=S_4\times \mathbb{Z}/7\mathbb{Z}$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.


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