# Show that a group of order $1755=3^3\times 5\times 13$ must have a normal Sylow $13$ subgroup or a normal Sylow $5$ subgroup.

Question: Show that a group of order $$1755=3^3\times 5\times 13$$ must have a normal Sylow $$13$$ subgroup or a normal Sylow $$5$$ subgroup.

My thoughts: $$n_{13}\equiv 1\pmod{13}$$ and $$n_{13}\mid 135$$, so $$n_{13}=1$$ or $$27$$. If $$n_{13}=1$$, then we have a normal Sylow $$13$$ subgroup of $$G$$, and we're done. If $$n_{13}=27$$, then let $$Q\in{\rm Syl}_{13}(G)$$. As $$G$$ acts on $$Q$$ by conjugation, we induce a homomorphism $$\phi:G\rightarrow S_{27}$$, where $$|G:N_G(Q)|=27$$, thus $$|N_G(Q)|=5\times 13$$. Since $$13\neq 1\pmod{5}$$, we have that $$N_G(Q)$$ is cyclic thus abelian. Let $$R\in{\rm Syl}_5(N_G(Q))$$, thus $$R\trianglelefteq N_G(Q)$$. From here, I am not quite sure how to show that $$R$$ must then be normal in $$G$$.

Any help is greatly appreciated!

Thank you.

Here is a proof. First, as you noticed, $$n_{13}=27$$ so there are $$12\cdot 27=324$$ elements of order $$13$$ in $$G$$. Next $$n_5$$ divides $$351$$ and $$n_5\equiv 1 \pmod{5}$$, hence $$n_5=351$$. Then the number of elements of order $$5$$ is $$351\cdot 4=1404$$. The number of elements of order $$5$$ or $$13$$ is then $$1728$$.
There are then at most $$1755-1728=27$$ elements whose orders are powers of $$3$$, including $$1$$. Hence $$n_3=1$$ and the group $$G$$ has one normal Sylow 3-subgroup $$N$$. For every $$5$$-Sylow subgroup $$S$$, $$SN/N$$ is a $$5$$-Sylow subgroup of $$G/N$$. If $$S'\ne S$$ is another Sylow $$5$$-subgroup, then $$SN/N\ne S'N/N$$. So $$G/N$$ has $$1404$$ elements of order $$5$$, a contradiction because $$|G/N|<1404$$.
• $S\neq S'$ does not automatically imply that $SN\neq S'N$. However, in the present case it does, because $SN$ has order $135$ and hence only a single Sylow $5$-subgroup. Commented Dec 29, 2021 at 8:39
• @JyrkiLahtonen: There are several other gaps in the answer which can be filled as well. For example, $SN/N$ is not necessarily a Sylow subgroup of $G/N$ but in this case it is. Commented Dec 29, 2021 at 8:46
• Couldn't we derive an earlier contradiction right after showing $SN/N \neq S'N /N$? Because then we would have shown that $G/N$ has more than one Sylow-5 subgroup, i.e. $n_5(G/N) > 1$ and yet Sylow's Third forces $n_5(G/N) = 1$ (since $|G/N| = 5 \cdot 13$). I'm not following how $G/N$ has $1404$ elements of order $5$. Commented Dec 28, 2023 at 14:57