# $G^F$ conjugacy class of $F$-stable maximal tori, in an algebraic group $G$ defined over $\mathbb{F}_{q}$.

Let $$G$$ be an affine algebraic group over $$k=\bar{\mathbb{F}_{p}}$$. Let $$q$$ be a power of $$p$$, and assume that $$G$$ is defined over $$\mathbb{F}_q$$. Let $$\mathcal{T}$$, be the collection of all maximal torus of $$G$$. Let $$F$$ be the Frobenius map from $$G$$ onto $$G$$.

Clearly $$G$$ acts on $$\mathcal{T}$$, by conjugation and this action is transitive. Further if we assume that $$G$$ is connected, by standard theory in "Finite groups of Lie type", $$G^{F}$$, acts on $$\mathcal{T}^{F}$$ by conjugation,where $$G^{F}$$ denote the set of $$F$$-rational points of $$G$$, which is the finite algebraic group associated to $$G$$, and $$\mathcal{T}^{F}$$, denote the collection of $$F-$$ stable maximal tori of $$G$$,that is,

$$\mathcal{T}^{F}=\{ T\in \mathcal{T} | F(T)=T \}.$$

We know, $$G^F$$ need not act transitively on $$\mathcal{T}^F$$. For each $$T\in \mathcal{T}^F$$, Let $$T^F$$, denote the $$F-$$ rational points in $$T$$. Now, let $$\mathcal{T}^{[F]}= \{ T^F | T\in \mathcal{T}^{F} \}.$$

It is clear that $$G^F$$ acts on $$\mathcal{T}^{[F]}$$, by conjugation. Also, if $$T_1, T_2 \in \mathcal{T}^F$$ are in same orbit under $$G^F$$ action then $$T_1^{F},T_2^{F}$$ are also in same orbit under $$G^F$$ action.

My question is Is it true other way around, that is, if $$T_1^{F}, T_2^{F}$$, are $$G^F$$-conjugate, then $$T_1,T_2$$ are $$G^F$$ conjugate. It seems to be that it is true. But, I couldn't come up with any proof.

I would appreciate any kind of help. Thank you!