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I am struggling with solving the following exercise in T.A springer's linear algebraic groups.

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I should add that in the context of the exercise $G$ is connected and solvable.

Kenta S
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roy yanai
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1 Answers1

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I think you can do this :

First let's show that $Z_G(H_s) \subseteq N_G(H)$. If $t \in Z_G(H_s)$, consider the group $K$ generated by $H_u$ and $tH_ut^{-1}$. The group $K$ is unipotent (this is a subgroup of $G_u$). Thanks to Question (4) we must have $K=H_u$, otherwise $H_u \subsetneq N_K(H_u) $ and $N_K(H_u)$ would be in $N_G(H)$ (because also $t \in Z_G(H_s)$) but not in $H$. Thus $tH_ut^{-1}=H_u$, that is $t \in N_G(H)$.

Then using 6.3.6(ii), $H \subseteq Z_G(H_s) \subseteq N_G(H)=H$, so $H=Z_G(H_s)$.

Finally let $T$ be a maximal torus of $G$ containing $H_s$ (this is 6.3.6(i)). We have $T \subseteq Z_G(H_s) \subseteq N_G(H)$, so $T \subseteq N_G(H)=H$ and $T=H_s$.