Question #158272

— 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 ...


1
Expert's answer
2021-01-26T11:57:14-0500

Let m1m_1 and m2m_2 belongs to the same orbit, that is m2=gm1m_2=g\circ m_1 for some element gg of a group GG. Let us show that St(m2)=gSt(m1)g1St(m_2)=gSt(m_1)g^{-1}.


Let hSt(m2)h\in St(m_2). Then hm2=m2h\circ m_2=m_2, and thus h(gm1)=gm1h\circ (g\circ m_1)=g\circ m_1. It follows from definition of action that (hg)m1=gm1(hg)\circ m_1=g\circ m_1, and therefore, g1((hg)m1)=g1(gm1)g^{-1}\circ((hg)\circ m_1)=g^{-1}\circ(g\circ m_1). Consequently, (g1hg)m1=(g1g)m1=em1=m1(g^{-1}hg)\circ m_1=(g^{-1}g)\circ m_1=e\circ m_1=m_1. We conclude that g1hgSt(m1)g^{-1}hg\in St(m_1), and thus hgSt(m1)g1h\in gSt(m_1)g^{-1}. It follows that St(m2)gSt(m1)g1St(m_2)\subset gSt(m_1)g^{-1}.


On the other hand, let hgSt(m1)g1h\in gSt(m_1)g^{-1}. Then h=gkg1h=gkg^{-1}, where kSt(m1)k\in St(m_1), that is km1=m1k\circ m_1=m_1. It follows that hm2=(gkg1)(gm1)=(gkg1g)m1=(gk)m1=g(km1)=gm1=m2h\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 hSt(m2)h\in St(m_2), and therefore, gSt(m1)g1St(m2)gSt(m_1)g^{-1}\subset St(m_2).


We concude that gSt(m1)g1=St(m2)gSt(m_1)g^{-1}= St(m_2), that is the stabilizers St(m1)St(m_1) and St(m2)St(m_2) are conjugate to each other.



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