Let $H$ and $K$ be two conjugate subgroups in $G.$ Show that $N_G(H)$ and $N_G(K)$ are conjugates.

Question

Let $H$ and $K$ be two conjugate subgroups in $G.$ Show that $N_G(H)$ and $N_G(K)$ are conjugates

I thought of using the second part of Sylow's Theorem that says; In particular, any two Sylow $p$-subgroups of $G$ are conjugate in $G$.

But I'm not sure if I have to prove that $H$ and $K$ are $p$-subgroups

Answer 1

Since $H$ and $K$ are conjugate in $G$, there exists a $g\in G$ such that $H=gKg^{-1}$.

Consider $$N_G(K)=\{x\in G\mid xK=Kx\}.$$

Observe that

$$\begin{align} gN_G(K)g^{-1}&=g\{x\in G\mid xK=Kx\} g^{-1}\\ &=\{gxg^{-1}\in G\mid xK=Kx\}\\ &=\{y=gxg^{-1}\in G\mid xKx^{-1}=K\}\\ &=\{y\in G\mid g^{-1}ygKg^{-1}y^{-1}g=K\}\\ &=\{y\in G\mid g^{-1}yHy^{-1}g=K\}\\ &=\{y\in G\mid yHy^{-1}=gKg^{-1}=H\}\\ &=\{y\in G\mid yH=Hy\}\\ &=N_G(H). \end{align}$$

Hence they are conjugate in $G$.

Answer 2

Hint: try to prove that $N_G(H^g)=N_G(H)^g$ for any subgroup $H$ of $G$ and element $g \in G$.

Answer 3

If $H$ and $K$ are conjugate in $G$, then they belong to the same orbit of the action of $G$ by conjugation on the set of the subgroups of $G$, and thence their stabilizers are conjugate. But such stabilizers are precisely $N_G(H)$ and $N_G(K)$, respectively.