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(Edit: the previous version of this answer was not correct, but I leave this here as a remark)

Let $\sigma: G \rightarrow G$ be the Frobenius map corresponding to the field automorphism $x \mapsto x^q$.

The set ofTo show that there exists a regular semisimple elements is invariant under $\sigma$. (Ifelement in $g \in G$ is semisimple$G_{\sigma} = G(q)$, then $g$by the Lang-Steinberg theorem it would be enough to show that there is a $\sigma$-invariant class of regular semisimple if and only if $C_G(g)^\circ$ is a toruselements.)

Thus it follows from See 2.7(a) in "Conjugacy Classes" by Springer and Steinberg (in Lecture Notes in Math 131) that there exists a regular semisimple element fixed by $\sigma$.

The proof of 2.7(a) does not need machinery other than the Lang–Steinberg theorem.

Let $\sigma: G \rightarrow G$ be the Frobenius map corresponding to the field automorphism $x \mapsto x^q$.

The set of regular semisimple elements is invariant under $\sigma$. (If $g \in G$ is semisimple, then $g$ is regular semisimple if and only if $C_G(g)^\circ$ is a torus.)

Thus it follows from 2.7(a) in "Conjugacy Classes" by Springer and Steinberg (in Lecture Notes in Math 131) that there exists a regular semisimple element fixed by $\sigma$.

The proof of 2.7(a) does not need machinery other than the Lang–Steinberg theorem.

(Edit: the previous version of this answer was not correct, but I leave this here as a remark)

Let $\sigma: G \rightarrow G$ be the Frobenius map corresponding to the field automorphism $x \mapsto x^q$.

To show that there exists a regular semisimple element in $G_{\sigma} = G(q)$, by the Lang-Steinberg theorem it would be enough to show that there is a $\sigma$-invariant class of regular semisimple elements. See 2.7(a) in "Conjugacy Classes" by Springer and Steinberg (in Lecture Notes in Math 131).

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Let $\sigma: G \rightarrow G$ be the Frobenius map corresponding to the field automorphism $x \mapsto x^q$.

The set of regular semisimple elements is invariant under $\sigma$. (If $g \in G$ is semisimple, then $g$ is regular semisimple if and only if $C_G(g)^\circ$ is a torus).)

Thus it follows from 2.7(a) in "Conjugacy Classes""Conjugacy Classes" by Springer and Steinberg (in Lecture Notes in Math 131 Lecture Notes in Math 131) that there exists a regular semisimple element fixed by $\sigma$.

The proof of 2.7(a) does not need machinery other than the Lang-SteinbergLang–Steinberg theorem.

Let $\sigma: G \rightarrow G$ be the Frobenius map corresponding to the field automorphism $x \mapsto x^q$.

The set of regular semisimple elements is invariant under $\sigma$. (If $g \in G$ is semisimple, then $g$ is regular semisimple if and only if $C_G(g)^\circ$ is a torus).

Thus it follows from 2.7(a) in "Conjugacy Classes" by Springer and Steinberg (in Lecture Notes in Math 131) that there exists a regular semisimple element fixed by $\sigma$.

The proof of 2.7(a) does not need machinery other than the Lang-Steinberg theorem.

Let $\sigma: G \rightarrow G$ be the Frobenius map corresponding to the field automorphism $x \mapsto x^q$.

The set of regular semisimple elements is invariant under $\sigma$. (If $g \in G$ is semisimple, then $g$ is regular semisimple if and only if $C_G(g)^\circ$ is a torus.)

Thus it follows from 2.7(a) in "Conjugacy Classes" by Springer and Steinberg (in Lecture Notes in Math 131) that there exists a regular semisimple element fixed by $\sigma$.

The proof of 2.7(a) does not need machinery other than the Lang–Steinberg theorem.

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Let $\sigma: G \rightarrow G$ be the Frobenius map corresponding to the field automorphism $x \mapsto x^q$.

The set of regular semisimple elements is invariant under $\sigma$. (If $g \in G$ is semisimple, then $g$ is regular semisimple if and only if $C_G(g)^\circ$ is a torus).

Thus it follows from 2.7(a) in "Conjugacy Classes" by Springer and Steinberg (in Lecture Notes in Math 131) that there exists a regular semisimple element fixed by $\sigma$.

The proof of 2.7(a) does not need machinery other than the Lang-Steinberg theorem.