# Counting the subgroups of order $6$ in a group of order $42$ which has a subgroup of order $6$

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 $$G$$.

• 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 $6$) – Derek Holt Nov 23 '18 at 11:58
• @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 $\frac{|G|}{|N_G(H) \cap K|}$"? – Math enthusiast Nov 23 '18 at 12:01
• 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 $6$. – Derek Holt Nov 23 '18 at 12:11
• 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 $N_G(H) = H$? – Math enthusiast Nov 23 '18 at 12:18
• Wait a moment, I think that I understood why $N_G(H) = H$. I will post my explanation soon – Math enthusiast Nov 23 '18 at 12:24

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 $$[G : N_G(H)] = \frac{|G|}{|N_G(H)|} = \frac{42}{6} = 7$$.