Dimension of normalizer of closed connected subgroup

Question

Let $G$ be unipotent and let $H$ be proper closed, connected subgroup. Show that $\operatorname{dim}(N_G(H)) > \operatorname{dim} H$.

We know that $H \triangleleft Z_G(H) \triangleleft N_G(H)$, therefore I want to prove that one of the inclusion is proper.

I think the latter inclusion is proper, but I have no idea how to prove.

Any help will be appreciate..

Answer 1

Let $L_G, L_H,L_N$ be the respective Lie algebras of $G, H$ and $N_G(H)$. Consider the action of $H$ on $L_G/L_H$ by $u_x(y)=p([x,y])$ where $p:L_G\rightarrow L_G/L_H$ is the quotient map, the theorem of Engel implies the existence of $x_0\neq 0\in L_G$ such that $u_x(x_0)=0$ for every $x\in L_H$, we deduce that $x_0\in L_N$ and $dim(N_G(H))>dim(H)$.

Answer 2

1. We should assume that $G$ is connected, otherwise we can let $H = G^{\circ} \Rightarrow \operatorname{dim}G \geq \operatorname{dim}N_GH > \operatorname{dim}H = \operatorname{dim}G$, contradiction.

2. Following is my proof, may have some mistakes.

Let $Z = C(G)^{\circ}$, i.e., the identity component of $C(G)$, then $Z$ is closed in $C(G)$, so $Z$ is closed in $G$ since $C(G)$ is closed in $G$.

Since $G$ is unipotent algebraic group, i.e. $G = G_u$, $G$ is solvable group. And by [Lemma 6.3.4, , T. A. Springer, Linear algebraic groups.], $\operatorname{dim}C(G)\geq1$.

If $Z \nsubseteq H$, then $\operatorname{dim}Z \geq 1$ and $|ZH/H| \geq 2$. But $ZH/H$ is connected since $ZH$ is connected, so $\operatorname{dim}(ZH) - \operatorname{dim}(H) = \operatorname{dim}(ZH/H) > 0$. Note that $ZH \subset N_G(H)$, so in this case we have $\operatorname{dim}N_G(H) \geq \operatorname{dim}ZH > \operatorname{dim}H \Rightarrow \operatorname{dim}N_G(H) > \operatorname{dim}H$.

Now we prove this by induction on $\operatorname{dim}G$. If $\operatorname{dim}G = 1$, then $G$ is commutative by [Prop 3.1.3, , T. A. Springer, Linear algebraic groups.], so $Z = G^{\circ} = G$, thus $Z \nsubseteq H$. By the above proof, we konw $\operatorname{dim}(N_G(H)) > \operatorname{dim}H$.

Suppose this is right for all $G$ satisfies $\operatorname{dim}G \leq n-1$. Consider this case $\operatorname{dim}G = n$. By the above proof, we may assume $Z \subseteq H$, then $G/Z$ is a unipotent connected group ($G$ is unipotent) and $$ \operatorname{dim} (G/Z) = \operatorname{dim}G - \operatorname{dim}Z = \operatorname{dim}G - \operatorname{dim}C(G) < \operatorname{dim}G. $$ Also, $H/Z$ is a proper connected closed group of $G/Z$. By induction, we have $$ \operatorname{dim}N_{G/Z}(H/Z) > \operatorname{dim}H/Z, $$ and $$ \operatorname{dim}N_GH - \operatorname{dim}Z = \operatorname{dim}N_GH/Z = \operatorname{dim}N_{G/Z}(H/Z) > \operatorname{dim}H/Z = \operatorname{dim} H- \operatorname{dim}Z. $$ Therefore, $\operatorname{dim}N_GH> \operatorname{dim} H$.