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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 $G$.

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  • $\begingroup$ Your answer is correct: if $G$ is nonabelian, then there are exactly $7$ cyclic subgroups of order $6$, and they are all conjugate. (There are actually three isomorphism classes of groups with this property, but they all have $7$ cyclic subgroups of order $6$) $\endgroup$ – Derek Holt Nov 23 '18 at 11:58
  • $\begingroup$ @DerekHolt, do you know what is the homomorphism which ensure my statement of that "the number of conjugates of $H$ with respect to $K$ is $\frac{|G|}{|N_G(H) \cap K|}$"? $\endgroup$ – Math enthusiast Nov 23 '18 at 12:01
  • $\begingroup$ That's not correct, because $|N_K(H)|=1$. The number of conjugates of $H$ is $|G/N_G(H)|$, but $N_K(H)=1$, so $N_G(H)= H$ has order $6$. $\endgroup$ – Derek Holt Nov 23 '18 at 12:11
  • $\begingroup$ Are you considering $N_K(H) = N_G(H) \cap K$, right? Why $|N_K(H)| = 1$? I thought $N_K(H)$ would have $6$ elements and how exactly did you conclude that $N_G(H) = H$? $\endgroup$ – Math enthusiast Nov 23 '18 at 12:18
  • $\begingroup$ Wait a moment, I think that I understood why $N_G(H) = H$. I will post my explanation soon $\endgroup$ – Math enthusiast Nov 23 '18 at 12:24
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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 $[G : N_G(H)] = \frac{|G|}{|N_G(H)|} = \frac{42}{6} = 7$.

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