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It is not clear to me in the first answer to this question in the last two lines why $P_{17}$ is abelian implies that $P_{17} \leq N_G(P_3)$ and what is the contradiction that we obtain that leads to $n_3 = 1.$

Here is the first answer there:

Hint: let's take a look at the number of groups in $Syl_{17}(G)$ and $Syl_3(G)$.

We know that $n_{17}\equiv1\text{ (mod 17)}\implies n_{17}\in \{1\not3,\not5,\not{15}\}\implies \exists!P_{17}\trianglelefteq G$.

While $n_3\equiv 1\text{ (mod 3)}\implies n_3\in\{1,\not5,\not17,85\}.$

If we now consider the action of conjugation of $G$ on its Sylow's groups we have that $$|Syl_{17}(G)|=|G:N_G(P_{17})|\implies|N_G(P_{17})|=|G|.$$ Furthermore $$\dfrac{|N_G(P_{17})|}{|C_G(P_{17})|}\bigg||\operatorname{Aut}(P_{17})|=\varphi(17)=16\implies\dfrac{3\cdot 5\cdot 17}{|C_G(P_{17})|}\big |2^4\iff|C_G(P_{17})|=|G|.$$ $P_{17}$ is abelian, so $P_{17}\le N_G(P_3)\implies17\big||N_G(P_{3})|$. We also have $\{1,85\}\ni n_3=|G:N_G(P_3)|$, so $n_3=1$. You can do something similar for $P_5$.

Could anyone help me to understand this please?

Shaun
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Emptymind
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  • As for your second question, $|N_G(P_3)|\ge 17\Longrightarrow [G:N_G(P_3)]\le 15<85$. – citadel Sep 08 '22 at 14:55
  • @Devo I did not get your point – Emptymind Sep 09 '22 at 08:51
  • If $17\mid |N_G(P_3)|$ (because $P_{17}\le N_G(P_3)$), then $|N_G(P_3)|$ is bigger than $17$, and hence the index of $N_G(P_3)$ in $G$ is less than $3\cdot 5=15$. But this index is equal to $n_3$, which is $1$ or $85$, so it must be $1$. Unless I misunderstood your point (and provided that the premises are true, which I took for granted but didn't verify.) – citadel Sep 09 '22 at 09:54

2 Answers2

2

I propose another solution.

We know that $P_{17}\lhd G$. Let $f:G\to \operatorname{Aut}(P_{17})$ be a homomorphism defined by the rule $f(g)(x)=gxg^{-1}$.

Since $ |\operatorname{Aut}(P_{17})|=16$ and $|G|$ is odd, it follows that $f(G)=1$. This means that $P_3<C_G(P_{17})$ and $P_5<C_G(P_{17})$. That's it.

Shaun
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kabenyuk
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  • I do not see why $f(G) = 1,$ could you please elaborate the theorem or lemma or any information you used to say this please? – Emptymind Sep 09 '22 at 08:29
  • By Lagrange's theorem $|f(G)|$ divides $|G|$ and $|f(G)|$ divides $|\operatorname{Aut}(P_{17})|=16$. Now consider that the numbers $|G|$ and $16$ are coprime. – kabenyuk Sep 09 '22 at 09:21
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Here's another way to see it, without using automorphism groups. Since there is a unique Sylow $17$-subgroup $P$ it is normal in $G$, and $G/P$ has order $15$. It is easy to see that groups of order $15$ are cyclic (they have a unique Sylow $3$- and $5$-subgroup) so there are exactly two elements of order $3$ in $G/P$. If $Q\leq G$ has order $3$ then $QP/P$ has order $3$ in $G/P$, so is the unique subgroup of order $3$ in $G/P$. Thus $QP$ is the unique subgroup of order $51$ in $G$ (so is normal in $G$), and is again cyclic. Thus $n_3=1$.

David A. Craven
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  • Similar: $G/P$ has order $15$, so by $5 \not\equiv 1 \mod{3}$ it has a normal subgroup of order $3$, so you have a normal subgroup $N$ of order $3 \cdot 17$ in $G$. Then by $17 \not\equiv 1 \mod{3}$ the normal subgroup $N$ has a unique $3$-Sylow, which is normal in $G$. – spin Sep 09 '22 at 07:12
  • I did not get what you are trying to do in your last 2 statements, could you clarify please? – Emptymind Sep 09 '22 at 08:47
  • @spin I did not get what theorem are you using to say this $5 \notequiv 1 mod 3$ imples that it has a normal subgroup of order $3,$ could you clarify please? – Emptymind Sep 09 '22 at 08:50
  • @Emptymind: That uses Sylow's theorem: the number of Sylow $p$-subgroups is $\equiv 1 \mod{p}$. And if there is only one Sylow $p$-subgroup, it must be normal. – spin Sep 09 '22 at 11:46
  • but you used "not congruent to" instead of "congruent to"@spin – Emptymind Sep 09 '22 at 16:40
  • @Emptymind: Yes, $5 \not\equiv 1 \mod{3}$ rules out the possibility that there would be $5$ Sylow $3$-subgroups in a group of order $15$. – spin Sep 10 '22 at 05:42