Answer to Question #158272 in Abstract Algebra for KV

Question

— If m1, m2 belongs to the same orbit then St(m1) and St(m2) are conjugate to each other if m2=p(g)m1 then St(m2)=gSt(m1) g ...

Answer 1

Let $m_1$ and $m_2$ belongs to the same orbit, that is $m_2=g\circ m_1$ for some element $g$ of a group $G$. Let us show that $St(m_2)=gSt(m_1)g^{-1}$.

Let $h\in St(m_2)$. Then $h\circ m_2=m_2$, and thus $h\circ (g\circ m_1)=g\circ m_1$. It follows from definition of action that $(hg)\circ m_1=g\circ m_1$, and therefore, $g^{-1}\circ((hg)\circ m_1)=g^{-1}\circ(g\circ m_1)$. Consequently, $(g^{-1}hg)\circ m_1=(g^{-1}g)\circ m_1=e\circ m_1=m_1$. We conclude that $g^{-1}hg\in St(m_1)$, and thus $h\in gSt(m_1)g^{-1}$. It follows that $St(m_2)\subset gSt(m_1)g^{-1}$.

On the other hand, let $h\in gSt(m_1)g^{-1}$. Then $h=gkg^{-1}$, where $k\in St(m_1)$, that is $k\circ m_1=m_1$. It follows that $h\circ m_2=(gkg^{-1})\circ(g\circ m_1)=(gkg^{-1}g)\circ m_1=(gk)\circ m_1=g\circ(k\circ m_1)=g\circ m_1=m_2$ . We conclude that $h\in St(m_2)$, and therefore, $gSt(m_1)g^{-1}\subset St(m_2)$.

We concude that $gSt(m_1)g^{-1}= St(m_2)$, that is the stabilizers $St(m_1)$ and $St(m_2)$ are conjugate to each other.