$|G|=105$, 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.
Best Answer
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 $|\langle xy\rangle|=15$.