Answer to Question #155733 in Abstract Algebra for KV

Question #155733

— 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
2021-01-17T17:52:04-0500

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}".


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