Prove PGL(V) Acts 2-Transitively on P(V) Projective Space

gankutsuou7
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Dim(V)>1.Prove that PGL(V) acts two transitively but not 3 transitively on P(V) projective space.
 
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Idea: Any two non-colinear vectors form a basis. Three vectors form a basis only when they are linearrly independent. You can't map linearly independent vectors to linearly dependent vectors by a GL(V) transformation.
 
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The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...

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