2

$|G|=105$, Show that is $P_3$ is a Sylow-3-subgroup then $5||N_G(P_3)|$

This question is given as an exercise here. I am having a hard time seeing it. From my Sylow's Theorem training I know that the number of Sylow-3-subgroups is the index in $G$ of the normalizer $N_G(P_3)$.

So I am thinking that we must first find the number of Sylow-3-subgroups, denoted by $n_3$. Then $n_3 \equiv 1 \mod 3$ and $n_3|35$ so $n_3 \in \{1,7\}$. If $n_3 = 1$ then $|N_G(P_3)|=105$, or if $n_3 = 7$ then $|N_G(P_3)|=15$. And 5 divides both of these possibilities. Is this the correct line of reasoning?

The following conclusion that $N_G(P_3)$ must contain a subgroup of order 15 is also escaping me. Of course, if $|N_G(P_3)|=15$, then we are done. Assuming the above is true, and the case that if $|N_G(P_3)|=105$, how can we conclude then that there exists a subgroup of order 15? I cannot assume that $G$ is abelian.

jeffery_the_wind
  • 1,089
  • Let $G$ be a finite group, $H$ be a normal subgroup and $K$ be a subgroup. The set $HK={hk | h\in H, k\in K}$ is in fact a subgroup of $G$ and if $gcd(|H|, |K|)=1$, you can conclude that $|HK|=|H||K|$. If $n_3=1$ or $n_5=1$, we are done. If $n_3 = 7, n_5 = 21$, $n_7=1$ and we directly get a subgroup of order 21. (This proof may not satisfy you, since the existence of subgroup of order 15 is not clear...) – nessy May 18 '20 at 15:33
  • 1
    Oh, if $N_G (P_3)=G$, $P_3$ is a normal subgroup, so $n_3 =1$ and we're done. – nessy May 18 '20 at 15:44
  • @nessy these 2 comments together make the best proof for me so far, thanks you! – jeffery_the_wind May 18 '20 at 23:05

1 Answers1

2

Let $x\in P_3$, let $y\in G$ have order $5$, and assume that $|N_G(P_3)|=105$, so that $y^{-1}xy\in P_3$. Clearly $xy\ne 1_G$, so $xy=yx$, or $xy=yx^2$.

If $xy=yx$, then $|\langle xy\rangle|=15$, and we’re done. Otherwise,

$$(xy)^5=yx^2(xy)^4=y^2(xy)^3=y^3x^2(xy)^2=y^4xy=x^2\;,$$

so $(xy)^{15}=(x^2)^3=1_G$, and again $|\langle xy\rangle|=15$.

Brian M. Scott
  • 616,228