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This question is connected to an extreme case of the odd analogue of Glauberman´s $Z^{\ast}$-theorem. This theorem asserts that if a finite group $G$ has no non-identity normal subgroup of order coprime to the prime $p,$ and $u$ is an element of order $p$ of $G$ which commutes with none of its other $G$-conjugates, then $u \in Z(G).$ This theorem was proved (by Glauberman) without CFSG for $p =2,$ but as far as I know all proofs to date for odd $p$ use CFSG.

Now suppose that $G$ is a doubly transitive permutation group with $F(G) =1,$ and that the point stabilizer $G_{x}$ has a central element $u$ of prime order $p$ (which will certainly happen if $Z(G_{x}) \neq 1).$ Then $G_{x} = C_{G}(u)$ and the permutation action is equivalent to that of $G$ acting by conjugation on the conjugates of $u.$

Now $C_{G}(u)$ permutes the conjugates of $u$ which commute with $u.$ By the double transitivity of the permutation action, we see that either all conjugates of $u$ commute with $u,$ or else no conjugate of $u$ other than $u$ itself commutes with $u.$ In the former case, $u \in F(G),$ which is excluded by hypothesis. Hence $u$ commutes with none of its other conjugates. By the general $Z^{\ast}$-theorem, we either have $u \in Z(G)$ or else $O_{p^{\prime}}(G) \neq 1.$ The former case is excluded as $F(G) = 1.$

Suppose then that $O_{p^{\prime}}(G) \neq 1.$ Then the image of $u$ is central in $G/O_{p^{\prime}}(G)$ and a Frattini argument yields $G = O_{p^{\prime}}(G)C_{G}(u).$

Since $u \not \in Z(G),$ there is a prime $q \neq p$ such that $u$ normalizes, but does not centralize, a Sylow $q$-subgroup, $Q$ say, of $O_{p^{\prime}}(G)$. Hence there is a $Q$-conjugate $v$ of $u$ such that $\langle u,v \rangle$ is a $\{p,q\}$-group. By the transitivity of $C_{G}(u)$ on the other conjugates of $u,$ it follows that $\langle u,w \rangle$ is a $\{p,q\}$-group for every conjugate $w$ of $u.$ It then follows that $[O_{p^{\prime}}(G),u] \lhd G$ is a non-trivial $q$-group, and that $F(G) \neq 1,$ contrary to hypothesis.

Later edit: I noticed that this also follows from Theorem D of :Guralnick, Robert M.; Robinson, Geoffrey R. On extensions of the Baer-Suzuki theorem. Israel J. Math. 82 (1993), no. 1-3, 281–297. One needs to use an earlier theorem of E. Shult. An outline proof is as follows:

We have a doubly transitive finite group $G$ with non-Abelian simple socle. Suppose that $H$ is a point stabilizer, and that $x$ is an element of prime order $p$ in $Z(H).$ Then $G$ acts doubly transitively by conjugation on the conjugates of $x$, so $C_{G}(x)$ is transitive on the remaining conjugates of $x$. There can be no other conjugate $y$ of $x$ which commutes with $x$, since if there were, all conjugates of $x$ would commute with $x$, contrary to the simplicity of $G$. By a theorem of E. Shult, there must be a $p^{\prime}$-subgroup $T$ of $G$ which is normalized, but not centralized. Hence for some $t \in T, x^{-1}x^{t}$ is a non-identity $p$-regular element. Since $C_{G}(x)$ is transitive on the remaining conjugates of $x$, we see that $x^{-1}x^{g}$ is $p$-regular for all $g \in G$. By Theorem D of the Guralnick-Robinson paper, and the argument above ( Theorem D gives $G = O_{p'}(G)C_{G}(x)$, and the argument above gives $[G,x] = [O_{p'}(G),x]$ is a $q$-group for some prime $q$, contrary to the fact that $G$ has non-Abelian simple socle).

This question is connected to an extreme case of the odd analogue of Glauberman´s $Z^{\ast}$-theorem. This theorem asserts that if a finite group $G$ has no non-identity normal subgroup of order coprime to the prime $p,$ and $u$ is an element of order $p$ of $G$ which commutes with none of its other $G$-conjugates, then $u \in Z(G).$ This theorem was proved (by Glauberman) without CFSG for $p =2,$ but as far as I know all proofs to date for odd $p$ use CFSG.

Now suppose that $G$ is a doubly transitive permutation group with $F(G) =1,$ and that the point stabilizer $G_{x}$ has a central element $u$ of prime order $p$ (which will certainly happen if $Z(G_{x}) \neq 1).$ Then $G_{x} = C_{G}(u)$ and the permutation action is equivalent to that of $G$ acting by conjugation on the conjugates of $u.$

Now $C_{G}(u)$ permutes the conjugates of $u$ which commute with $u.$ By the double transitivity of the permutation action, we see that either all conjugates of $u$ commute with $u,$ or else no conjugate of $u$ other than $u$ itself commutes with $u.$ In the former case, $u \in F(G),$ which is excluded by hypothesis. Hence $u$ commutes with none of its other conjugates. By the general $Z^{\ast}$-theorem, we either have $u \in Z(G)$ or else $O_{p^{\prime}}(G) \neq 1.$ The former case is excluded as $F(G) = 1.$

Suppose then that $O_{p^{\prime}}(G) \neq 1.$ Then the image of $u$ is central in $G/O_{p^{\prime}}(G)$ and a Frattini argument yields $G = O_{p^{\prime}}(G)C_{G}(u).$

Since $u \not \in Z(G),$ there is a prime $q \neq p$ such that $u$ normalizes, but does not centralize, a Sylow $q$-subgroup, $Q$ say, of $O_{p^{\prime}}(G)$. Hence there is a $Q$-conjugate $v$ of $u$ such that $\langle u,v \rangle$ is a $\{p,q\}$-group. By the transitivity of $C_{G}(u)$ on the other conjugates of $u,$ it follows that $\langle u,w \rangle$ is a $\{p,q\}$-group for every conjugate $w$ of $u.$ It then follows that $[O_{p^{\prime}}(G),u] \lhd G$ is a non-trivial $q$-group, and that $F(G) \neq 1,$ contrary to hypothesis.

Later edit: I noticed that this also follows from Theorem D of :Guralnick, Robert M.; Robinson, Geoffrey R. On extensions of the Baer-Suzuki theorem. Israel J. Math. 82 (1993), no. 1-3, 281–297. One needs to use an earlier theorem of E. Shult. An outline proof is as follows:

We have a doubly transitive finite group $G$ with non-Abelian simple socle. Suppose that $H$ is a point stabilizer, and that $x$ is an element of prime order $p$ in $Z(H).$ Then $G$ acts doubly transitively by conjugation on the conjugates of $x$, so $C_{G}(x)$ is transitive on the remaining conjugates of $x$. There can be no other conjugate $y$ of $x$ which commutes with $x$, since if there were, all conjugates of $x$ would commute with $x$, contrary to the simplicity of the socle of $G$. By a theorem of E. Shult, there must be a $p^{\prime}$-subgroup $T$ of $G$ which is normalized, but not centralized. Hence for some $t \in T, x^{-1}x^{t}$ is a non-identity $p$-regular element. Since $C_{G}(x)$ is transitive on the remaining conjugates of $x$, we see that $x^{-1}x^{g}$ is $p$-regular for all $g \in G$. By Theorem D of the Guralnick-Robinson paper, and the argument above ( Theorem D gives $G = O_{p'}(G)C_{G}(x)$, and the argument above gives $[G,x] = [O_{p'}(G),x]$ is a $q$-group for some prime $q$, contrary to the fact that $G$ has non-Abelian simple socle).

This question is connected to an extreme case of the odd analogue of Glauberman´s $Z^{\ast}$-theorem. This theorem asserts that if a finite group $G$ has no non-identity normal subgroup of order coprime to the prime $p,$ and $u$ is an element of order $p$ of $G$ which commutes with none of its other $G$-conjugates, then $u \in Z(G).$ This theorem was proved (by Glauberman) without CFSG for $p =2,$ but as far as I know all proofs to date for odd $p$ use CFSG.

Now suppose that $G$ is a doubly transitive permutation group with $F(G) =1,$ and that the point stabilizer $G_{x}$ has a central element $u$ of prime order $p$ (which will certainly happen if $Z(G_{x}) \neq 1).$ Then $G_{x} = C_{G}(u)$ and the permutation action is equivalent to that of $G$ acting by conjugation on the conjugates of $u.$

Now $C_{G}(u)$ permutes the conjugates of $u$ which commute with $u.$ By the double transitivity of the permutation action, we see that either all conjugates of $u$ commute with $u,$ or else no conjugate of $u$ other than $u$ itself commutes with $u.$ In the former case, $u \in F(G),$ which is excluded by hypothesis. Hence $u$ commutes with none of its other conjugates. By the general $Z^{\ast}$-theorem, we either have $u \in Z(G)$ or else $O_{p^{\prime}}(G) \neq 1.$ The former case is excluded as $F(G) = 1.$

Suppose then that $O_{p^{\prime}}(G) \neq 1.$ Then the image of $u$ is central in $G/O_{p^{\prime}}(G)$ and a Frattini argument yields $G = O_{p^{\prime}}(G)C_{G}(u).$

Since $u \not \in Z(G),$ there is a prime $q \neq p$ such that $u$ normalizes, but does not centralize, a Sylow $q$-subgroup, $Q$ say, of $O_{p^{\prime}}(G)$. Hence there is a $Q$-conjugate $v$ of $u$ such that $\langle u,v \rangle$ is a $\{p,q\}$-group. By the transitivity of $C_{G}(u)$ on the other conjugates of $u,$ it follows that $\langle u,w \rangle$ is a $\{p,q\}$-group for every conjugate $w$ of $u.$ It then follows that $[O_{p^{\prime}}(G),u] \lhd G$ is a non-trivial $q$-group, and that $F(G) \neq 1,$ contrary to hypothesis.

Later edit: I noticed that this also follows from Theorem D of :Guralnick, Robert M.; Robinson, Geoffrey R. On extensions of the Baer-Suzuki theorem. Israel J. Math. 82 (1993), no. 1-3, 281–297. One needs to use an earlier theorem of E. Shult. An outline proof is as follows:

We have a doubly transitive finite simple group $G$. Suppose that $H$ is a point stabilizer, and that $x$ is an element of prime order $p$ in $Z(H).$ Then $G$ acts doubly transitively by conjugation on the conjugates of $x$, so $C_{G}(x)$ is transitive on the remaining conjugates of $x$. There can be no other conjugate $y$ of $x$ which commutes with $x$, since if there were, all conjugates of $x$ would commute with $x$, contrary to the simplicity of $G$. By a theorem of E. Shult, there must be a $p^{\prime}$-subgroup $T$ of $G$ which is normalized, but not centralized. Hence for some $t \in T, x^{-1}x^{t}$ is a non-identity $p$-regular element. Since $C_{G}(x)$ is transitive on the remaining conjugates of $x$, we see that $x^{-1}x^{g}$ is $p$-regular for all $g \in G$. By Theorem D of the Guralnick-Robinson paper, we conclude that $x \in Z(G)$ as $G$ is simple of order divisible by $p,$ a contradiction.

This question is connected to an extreme case of the odd analogue of Glauberman´s $Z^{\ast}$-theorem. This theorem asserts that if a finite group $G$ has no non-identity normal subgroup of order coprime to the prime $p,$ and $u$ is an element of order $p$ of $G$ which commutes with none of its other $G$-conjugates, then $u \in Z(G).$ This theorem was proved (by Glauberman) without CFSG for $p =2,$ but as far as I know all proofs to date for odd $p$ use CFSG.

Now suppose that $G$ is a doubly transitive permutation group with $F(G) =1,$ and that the point stabilizer $G_{x}$ has a central element $u$ of prime order $p$ (which will certainly happen if $Z(G_{x}) \neq 1).$ Then $G_{x} = C_{G}(u)$ and the permutation action is equivalent to that of $G$ acting by conjugation on the conjugates of $u.$

Now $C_{G}(u)$ permutes the conjugates of $u$ which commute with $u.$ By the double transitivity of the permutation action, we see that either all conjugates of $u$ commute with $u,$ or else no conjugate of $u$ other than $u$ itself commutes with $u.$ In the former case, $u \in F(G),$ which is excluded by hypothesis. Hence $u$ commutes with none of its other conjugates. By the general $Z^{\ast}$-theorem, we either have $u \in Z(G)$ or else $O_{p^{\prime}}(G) \neq 1.$ The former case is excluded as $F(G) = 1.$

Suppose then that $O_{p^{\prime}}(G) \neq 1.$ Then the image of $u$ is central in $G/O_{p^{\prime}}(G)$ and a Frattini argument yields $G = O_{p^{\prime}}(G)C_{G}(u).$

Since $u \not \in Z(G),$ there is a prime $q \neq p$ such that $u$ normalizes, but does not centralize, a Sylow $q$-subgroup, $Q$ say, of $O_{p^{\prime}}(G)$. Hence there is a $Q$-conjugate $v$ of $u$ such that $\langle u,v \rangle$ is a $\{p,q\}$-group. By the transitivity of $C_{G}(u)$ on the other conjugates of $u,$ it follows that $\langle u,w \rangle$ is a $\{p,q\}$-group for every conjugate $w$ of $u.$ It then follows that $[O_{p^{\prime}}(G),u] \lhd G$ is a non-trivial $q$-group, and that $F(G) \neq 1,$ contrary to hypothesis.

Later edit: I noticed that this also follows from Theorem D of :Guralnick, Robert M.; Robinson, Geoffrey R. On extensions of the Baer-Suzuki theorem. Israel J. Math. 82 (1993), no. 1-3, 281–297. One needs to use an earlier theorem of E. Shult. An outline proof is as follows:

We have a doubly transitive finite group $G$ with non-Abelian simple socle. Suppose that $H$ is a point stabilizer, and that $x$ is an element of prime order $p$ in $Z(H).$ Then $G$ acts doubly transitively by conjugation on the conjugates of $x$, so $C_{G}(x)$ is transitive on the remaining conjugates of $x$. There can be no other conjugate $y$ of $x$ which commutes with $x$, since if there were, all conjugates of $x$ would commute with $x$, contrary to the simplicity of $G$. By a theorem of E. Shult, there must be a $p^{\prime}$-subgroup $T$ of $G$ which is normalized, but not centralized. Hence for some $t \in T, x^{-1}x^{t}$ is a non-identity $p$-regular element. Since $C_{G}(x)$ is transitive on the remaining conjugates of $x$, we see that $x^{-1}x^{g}$ is $p$-regular for all $g \in G$. By Theorem D of the Guralnick-Robinson paper, and the argument above ( Theorem D gives $G = O_{p'}(G)C_{G}(x)$, and the argument above gives $[G,x] = [O_{p'}(G),x]$ is a $q$-group for some prime $q$, contrary to the fact that $G$ has non-Abelian simple socle).

This question is connected to an extreme case of the odd analogue of Glauberman´s $Z^{\ast}$-theorem. This theorem asserts that if a finite group $G$ has no non-identity normal subgroup of order coprime to the prime $p,$ and $u$ is an element of order $p$ of $G$ which commutes with none of its other $G$-conjugates, then $u \in Z(G).$ This theorem was proved (by Glauberman) without CFSG for $p =2,$ but as far as I know all proofs to date for odd $p$ use CFSG.

Now suppose that $G$ is a doubly transitive permutation group with $F(G) =1,$ and that the point stabilizer $G_{x}$ has a central element $u$ of prime order $p$ (which will certainly happen if $Z(G_{x}) \neq 1).$ Then $G_{x} = C_{G}(u)$ and the permutation action is equivalent to that of $G$ acting by conjugation on the conjugates of $u.$

Now $C_{G}(u)$ permutes the conjugates of $u$ which commute with $u.$ By the double transitivity of the permutation action, we see that either all conjugates of $u$ commute with $u,$ or else no conjugate of $u$ other than $u$ itself commutes with $u.$ In the former case, $u \in F(G),$ which is excluded by hypothesis. Hence $u$ commutes with none of its other conjugates. By the general $Z^{\ast}$-theorem, we either have $u \in Z(G)$ or else $O_{p^{\prime}}(G) \neq 1.$ The former case is excluded as $F(G) = 1.$

Suppose then that $O_{p^{\prime}}(G) \neq 1.$ Then the image of $u$ is central in $G/O_{p^{\prime}}(G)$ and a Frattini argument yields $G = O_{p^{\prime}}(G)C_{G}(u).$

Since $u \not \in Z(G),$ there is a prime $q \neq p$ such that $u$ normalizes, but does not centralize, a Sylow $q$-subgroup, $Q$ say, of $O_{p^{\prime}}(G)$. Hence there is a $Q$-conjugate $v$ of $u$ such that $\langle u,v \rangle$ is a $\{p,q\}$-group. By the transitivity of $C_{G}(u)$ on the other conjugates of $u,$ it follows that $\langle u,w \rangle$ is a $\{p,q\}$-group for every conjugate $w$ of $u.$ It then follows that $[O_{p^{\prime}}(G),u] \lhd G$ is a non-trivial $q$-group, and that $F(G) \neq 1,$ contrary to hypothesis.

This question is connected to an extreme case of the odd analogue of Glauberman´s $Z^{\ast}$-theorem. This theorem asserts that if a finite group $G$ has no non-identity normal subgroup of order coprime to the prime $p,$ and $u$ is an element of order $p$ of $G$ which commutes with none of its other $G$-conjugates, then $u \in Z(G).$ This theorem was proved (by Glauberman) without CFSG for $p =2,$ but as far as I know all proofs to date for odd $p$ use CFSG.

Now suppose that $G$ is a doubly transitive permutation group with $F(G) =1,$ and that the point stabilizer $G_{x}$ has a central element $u$ of prime order $p$ (which will certainly happen if $Z(G_{x}) \neq 1).$ Then $G_{x} = C_{G}(u)$ and the permutation action is equivalent to that of $G$ acting by conjugation on the conjugates of $u.$

Now $C_{G}(u)$ permutes the conjugates of $u$ which commute with $u.$ By the double transitivity of the permutation action, we see that either all conjugates of $u$ commute with $u,$ or else no conjugate of $u$ other than $u$ itself commutes with $u.$ In the former case, $u \in F(G),$ which is excluded by hypothesis. Hence $u$ commutes with none of its other conjugates. By the general $Z^{\ast}$-theorem, we either have $u \in Z(G)$ or else $O_{p^{\prime}}(G) \neq 1.$ The former case is excluded as $F(G) = 1.$

Suppose then that $O_{p^{\prime}}(G) \neq 1.$ Then the image of $u$ is central in $G/O_{p^{\prime}}(G)$ and a Frattini argument yields $G = O_{p^{\prime}}(G)C_{G}(u).$

Since $u \not \in Z(G),$ there is a prime $q \neq p$ such that $u$ normalizes, but does not centralize, a Sylow $q$-subgroup, $Q$ say, of $O_{p^{\prime}}(G)$. Hence there is a $Q$-conjugate $v$ of $u$ such that $\langle u,v \rangle$ is a $\{p,q\}$-group. By the transitivity of $C_{G}(u)$ on the other conjugates of $u,$ it follows that $\langle u,w \rangle$ is a $\{p,q\}$-group for every conjugate $w$ of $u.$ It then follows that $[O_{p^{\prime}}(G),u] \lhd G$ is a non-trivial $q$-group, and that $F(G) \neq 1,$ contrary to hypothesis.

Later edit: I noticed that this also follows from Theorem D of :Guralnick, Robert M.; Robinson, Geoffrey R. On extensions of the Baer-Suzuki theorem. Israel J. Math. 82 (1993), no. 1-3, 281–297. One needs to use an earlier theorem of E. Shult. An outline proof is as follows:

We have a doubly transitive finite simple group $G$. Suppose that $H$ is a point stabilizer, and that $x$ is an element of prime order $p$ in $Z(H).$ Then $G$ acts doubly transitively by conjugation on the conjugates of $x$, so $C_{G}(x)$ is transitive on the remaining conjugates of $x$. There can be no other conjugate $y$ of $x$ which commutes with $x$, since if there were, all conjugates of $x$ would commute with $x$, contrary to the simplicity of $G$. By a theorem of E. Shult, there must be a $p^{\prime}$-subgroup $T$ of $G$ which is normalized, but not centralized. Hence for some $t \in T, x^{-1}x^{t}$ is a non-identity $p$-regular element. Since $C_{G}(x)$ is transitive on the remaining conjugates of $x$, we see that $x^{-1}x^{g}$ is $p$-regular for all $g \in G$. By Theorem D of the Guralnick-Robinson paper, we conclude that $x \in Z(G)$ as $G$ is simple of order divisible by $p,$ a contradiction.