Let be $G$ a group of order $42$. Suppose $G$ has a subgroup of order $6$. Compute the number of conjugates of this subgroup in $G$.
This is what I thought:
Let be $H$ the subgroup of order $6$ and $K$ the only one $7$-subgroup of Sylow.
If $G$ is abelian, then there is an only one conjugate of $H$ in $G$. If $G$ is non abelian, we proceed as follow:
Acting $K$ on $H$ by conjugation,
$$|H| \equiv |N_G(H) \cap K| \ \text{mod} \ 7$$
by the theorem $4.1$ on page $18$ of this lecture notes, therefore $|N_G(H) \cap K| = 6$. On the one hand, the number of conjugates of $H$ with respect to $K$ is $\frac{|G|}{|N_G(H) \cap K|} = \frac{42}{6} = 7$. By the other hand, $hHh^{-1} = H$ for every $h \in H$, but $eHe^{-1}$ is a conjugate of $H$ with respect to $e \in K$, then $hHh^{-1} = eHe^{-1}$ for every $h \in H$, therefore $H$ has exactly $7$ conjugates in $G$. $\square$
I don't sure about this argument, because I stated "the number of conjugates of $H$ with respect to $K$ is $\frac{|G|}{|N_G(H) \cap K|}$". I know that it's true that the number of conjugates of $H$ in $G$ is $\frac{|G|}{|N_G(H)|}$, but I don't sure if what I stated it's true, I couldn't realize what action group I need to define to ensure that my statement it's true. I would like to know if what I do is correct and if it is how ensure that the number of conjugates of $H$ with respect to $K$ is $\frac{|G|}{|N_G(H) \cap K|}$. If this is not correct, I would like a hint in order to compute the number of conjugates of $H$ in $G$.
Best Answer
It was proved on the comments that if $G$ is non abelian and $H \leq G$ such that $|H| = 6$, then $N_G(H) = H$. Thus, the number of conjugates of $H$ is $[G : N_G(H)] = \frac{|G|}{|N_G(H)|} = \frac{42}{6} = 7$.