Answer to Question #155733 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

As $m_1,m_2$ are in the same orbit, $\exists g , m_2=\rho(g)m_1$. We can also note that $m_1=\rho(g^{-1})m_2$, so to prove that $Stab(m_2)= gStab(m_1)g^{-1}$ it is enough to prove that $gStab(m_1)g^{-1} \subset Stab(m_2)$ as the other inclusion can be obtained just by changing the roles of $m_1$ and $m_2$ and replacing $g$ by $g^{-1}$. Suppose $x\in Stab(m_1)$, now let's apply $\rho(gxg^{-1})$ to $m_2$ :

$\rho(gxg^{-1})m_2 = \rho(g)\rho(x)\rho(g^{-1})m_2 =\rho(g)\rho(x) m_1=\rho(g)m_1=m_2$. Therefore $gxg^{-1}\in Stab(m_2)$ and we have $gStab(m_1)g^{-1}\subset Stab(m_2)$. Now let's take an element $y \in Stab(m_2)$. To prove that $y\in gStab(m_1)g^{-1}$ we need to prove that $g^{-1}yg \in Stab(m_1)$. We verify it directly :

$\rho(g^{-1}yg) m_1 =\rho(g^{-1}) \rho(y)\rho(g)m_1 =\rho(g^{-1})\rho(y) m_2=\rho(g^{-1})m_2=m_1$. Thus $g^{-1}yg\in Stab(m_1)$, so $g^{-1}Stab(m_2)g \subset Stab(m_1); Stab(m_2)\subset gStab(m_1)g^{-1}$. Therefore we have $Stab(m_2)=gStab(m_1)g^{-1}$.