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How can a Frobenius map permute the roots of an algebraic group?

According to Carter (in Finite groups of Lie type), a root subgroup $X_{\alpha}$ is the 1-dimensional unipotenet subgroup giving rise to the root $\alpha$. The permutation $\rho$ on the roots, which is given by a Frobenius map $F: G \rightarrow G$ is defined to be $F(X_{\alpha}) = X_{\rho(\alpha)}$.

But I don't understand the permutation. For example, according to the figure on page 37, there is certain Frobenius map for a group of type $A_l$, $l \geq 2$, such that two of the simple roots are permuted.

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But I don't know how can this take place.

Now I am considering a special case. Let $K = \bar{\mathbb F_2}$, the algebraic closure of the field of two elements, and $G = SL(3,K)$, the group of $3 \times 3$ special linear group over $K$, which is of type $A_2$. If $F: G \rightarrow G$, $(a_{ij}) \mapsto (a_{ij}^2)$, and for two simple roots $\alpha, \beta$ of $G$, $X_{\alpha} =\left\{ \begin{pmatrix} 1 & a & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} | a \in K \right\}$, $X_{\beta} =\left\{ \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & b\\ 0 & 0 & 1 \end{pmatrix} | b \in K \right\}$. Then both $F(X_{\alpha})$ and $F(X_{\beta})$ are all themselves, with no permutation done. From this, I can see no hint as to the case when the permutation can be caused by a Frobenius map.

Is there any concerete example of this permutation? Any type of algebraic groups is OK. Thanks very much.

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