By using Sylow theorem, I can prove that $G$ has either $1$ Sylow $3$-subgroup or $4$ Sylow $3$-subgroup, but I don't know how to continue the proof.

  • 4
    $\begingroup$ Most questions of this type (when you first see them) can be solved using one of the following three methods. 1. Show that, for some prime $p$, $G$ must have a unique Sylow-p subgroup. 2. Show that if $G$ fails to have any normal Sylow subgroups, you end up accounting for more elements of Sylow subgroups then there are in your group. 3. Show that, if $G$ has a Sylow p subgroup, then $|G|$ does not divide $n_p(G)!$. This implies that $G$ is non-simple. Try method 3. $\endgroup$ Jun 22, 2014 at 4:11
  • $\begingroup$ math.stackexchange.com/questions/480558/… would be of some help i guess... $\endgroup$
    – user87543
    Jun 22, 2014 at 5:28
  • $\begingroup$ @AbrahamFrei-Pearson Sorry? What's $n_p(G)$ here means? I don't quite know. $\endgroup$
    – Jadim
    Jun 22, 2014 at 6:24
  • $\begingroup$ The number of distinct Sylow-p subgroups in $G$. If $P$ is a Sylow-p subgroup, then $n_p(G)=|G:N_{G}(P)|$. $\endgroup$ Jun 22, 2014 at 18:38

1 Answer 1


If there is a unique Sylow $3$-subgroup, it is nontrivial and normal, and you are done.

If there are $4$, then $G$ acts on these subgroups by conjugation, inducing a homomorphism $G\to S_4$. Since $|S_4|=24$, the kernel of this map cannot be trivial, otherwise $G$ would inject into $S_4$.

So it remains to show that the kernel is not all of $G$. Mouse over if you want to see the rest.

If it were, then every $g\in G$ normalizes each Sylow $3$-subgroup, implying each is normal, a contradiction. So the kernel is a nontrivial normal subgroup of $G$.

  • 1
    $\begingroup$ Thank you! I've understood it. $\endgroup$
    – Jadim
    Jun 22, 2014 at 16:20
  • 1
    $\begingroup$ That is a beautiful solution! I just had to comment $\endgroup$
    – iYOA
    Aug 10, 2018 at 4:23

You must log in to answer this question.

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