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2 sharply 2-trans

Suppose G=SL(2,C) and let H be the stabilizer of a line (so a Borel subgroup). The matrix $$\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ in G acts on projective space by $$z \mapsto \frac{az+b}{cz+d}$$ The stabilizer of ∞ is the matrices with c=0, a Borel subgroup. The stabilizer of both ∞ and 0 is the matrices with b=c=0, a maximal torus. In particular, a two point stabilizer is abelian (the intersection of two Borel subgroups), and a Borel subgroup is non-abelian. Hence they are not isomorphic.

This is just a connected version of Giuseppe's answer.

If you consider the Borel subgroup to be the Lie group itself, then you get an example where the intersection of the conjugates is trivial. If G is the group of 2×2 matrices with c=0 and d=1 acting on projective space, then the stabilizer of 0 has b=0, and the stabilizer of both 0 and 1 has a=1 and b=0. In particular, G=AGL(1) and H is a maximal torus, and the intersection of two conjugates of H is the identity. When the intersection is the identity, this is called being sharply two-transitive or having a regular stabilizer.

1

Suppose G=SL(2,C) and let H be the stabilizer of a line (so a Borel subgroup). The matrix $$\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ in G acts on projective space by $$z \mapsto \frac{az+b}{cz+d}$$ The stabilizer of ∞ is the matrices with c=0, a Borel subgroup. The stabilizer of both ∞ and 0 is the matrices with b=c=0, a maximal torus. In particular, a two point stabilizer is abelian (the intersection of two Borel subgroups), and a Borel subgroup is non-abelian. Hence they are not isomorphic.

This is just a connected version of Giuseppe's answer.