Can $1\ne H\cap H^g\lhd H$ happen if $G$ is a primitive permutation group with stabiliser $H$? Assume everything is finite.
Let $G$ be a primitive permutation group with point stabiliser $G_\alpha$ for some $\alpha$. For $\beta\ne\alpha$, by an arc stabiliser we mean $G_{\alpha\beta}=G_\alpha\cap G_\beta$ and an edge stabiliser we mean $G_{\{\alpha,\beta\}}$, the stabiliser of the set $\{\alpha,\beta\}$. Note that $G_{\{\alpha,\beta\}}\ge G_{\alpha\beta}$.
My question is: can an arc stabiliser $G_{\alpha\beta}\ne 1$ be normal in $G_\alpha$?
This is impossible if $G_{\{\alpha,\beta\}}> G_{\alpha\beta}$. In this case, there exists $g\in G$ such that $\alpha^g=\beta$ and $\beta^g=\alpha$. It follows that $g\in N_G(G_{\alpha\beta})$. Note that $G_\alpha$ is maximal in $G$ and $G_{\alpha\beta}$ cannot be normal in $G$ (otherwise $G_{\alpha\beta}$ will be in the kernel of this action), the normaliser $N_G(G_{\alpha\beta})=G_\alpha$. This gives a contradiction: $g\in G_\alpha$ but $\alpha^g=\beta\ne \alpha$.
I think there might exist an example in the case when $G_{\{\alpha,\beta\}}=G_{\alpha\beta}$. However, by a quick check with Magma there is no such primitive permutation group $G$ with $|G|\le 300$.

An equivalent statement is: Let $G$ be a group and $H$ a maximal and core-free subgroup of $G$. Is it possible that $1\ne H\cap H^g\lhd H$ for some $g\notin H$?
As shown in the comments there by Verret and Holt there is no such example for degree $\le 4095$. I also think the proof or an example, if any, will not be elementary (for example, applying O'Nan-Scott Theorem).

This was initially a MSE question I asked.
 A: Sorry, this is not an answer, but rather an application of standard techniques of local analysis to obtain substantial structural information about $H\cap H^g$, in case an example exists. The techniques are inventions of John Thompson, George Glauberman, and Helmut Bender.
Proposition: Let $G$ be a primitive permutation group on a set $\Omega$, and let $\alpha$ and $\beta$ be distinct points in $\Omega$. If $1\ne G_{\alpha\beta}\triangleleft G_\alpha$, then $G_{\alpha\beta}$ has the following properties:
(a) $F^*(G_{\alpha\beta})$ is either a $p$-group for some prime $p$, or the direct product of nonabelian simple groups.
(b) If $G_{\alpha\beta}$ is solvable, then $F^*(G_{\alpha\beta})$ is a $2$-group or a $3$-group.
(c) If $P$ is a Sylow subgroup of $G_{\alpha\beta}$, then no nontrivial characteristic subgroup of $P$ is normal in $G_{\alpha\beta}$.
(d) If $F^*(G_{\alpha\beta})$ is a $p$-group, $p>2$, then $G_{\alpha\beta}$ is not $p$-stable.
(e) (Uses Odd Order Theorem) $G_{\alpha\beta}$ has even order.
Proof: First, part (c). Suppose that $P$ is Sylow in $G_{\alpha\beta}$, $P_0\ne 1$ is a characteristic subgroup of $P$, and $P_0\triangleleft G_{\alpha\beta}$. By the Frattini argument $G_\alpha=G_{\alpha\beta}N_{G_\alpha}(P)\le N_{G_\alpha}(P_0)$ so $G_\alpha=N_G(P_0)$ by maximality of $G_\alpha$. Hence $N_{G_\beta}(P)\le N_{G_\beta}(P_0)=G_{\alpha\beta}$. Since $P$ is Sylow in $G_{\alpha\beta}$, it is Sylow in $G_\beta$ and hence also in $G_\alpha\cong G_\beta$. Now $G_\beta=G_\alpha^g$ for some $g\in G$, so $P^{gh}=P$ for some $h\in G_\beta$, by Sylow's theorem. Then $gh\in N_G(P)\le N_G(P_0)$ so $P_0\triangleleft G_\alpha^{gh}=G_\beta$. Then $P_0\triangleleft\langle G_\alpha,G_\beta\rangle=G$, so $P_0=1$ as $G_\alpha$ contains no nontrivial normal subgroup of $G$.
But $P_0\ne1$, contradicttion.
Next, part (a). If $F(G_{\alpha\beta})=1$, then $F^*(G_{\alpha\beta})=E(G_{\alpha\beta})$ and it is the direct product of nonabelian simple groups. So suppose $B=O_p(G_{\alpha\beta})\ne 1$ for some prime $p$. Then $N_G(B)=G_\alpha$. Let $A=O^p(F^*(G_{\alpha\beta}))$, the subgroup of $F^*(G_{\alpha\beta})$ generated by all its $p'$-elements. By the structure of generalized Fitting subgroups, $[A,B]=1$. If $A=1$, then $F^*(G_{\alpha\beta})=B$ and we are done, so assume that $A\ne 1$. Then $N_G(A)=G_\alpha$. Let $a\in A$ be an arbitrary $p'$-element and consider the action of $\langle a\rangle\times B$ on $C=O_p(G_\beta)\triangleleft G_\beta$.
We have $[\langle a\rangle,C_C(B)]\le [A,G_\alpha]\cap[G_\beta,C]\le A\cap C\le O_p(A)$. But $A$ is the elementwise-commuting product of $q$-groups for various primes $q\ne p$, and quasisimple groups. Therefore $[\langle a\rangle, C_C(B)]\le Z(A)$ so $[\langle a\rangle,C_C(B),\langle a\rangle]=1$. As $a$ is a $p'$-element normalizing the $p$-group $C_C(B)$, a standard coprime action lemma implies that $[\langle a\rangle,C_C(B)]=1$.Then the Thompson $A\times B$ Lemma implies that $[\langle a\rangle,C]=1$.  But $a$ was an arbitrary   generator of $A$, so $[A,C]=1$. This implies in turn that $C\le N_G(A)=G_\alpha$. Since $C$ is a normal $p$-subgroup of $G_\beta$, $C\le O_p(G_{\alpha\beta})=B$. But $B\le O_p(G_\alpha)\cong O_p(G_\beta)=C$, so $B=C$. Since $B\triangleleft G_\alpha$ and $C\triangleleft G_\beta$, $B\triangleleft G$, a contradiction as $B\le G_{\alpha}$.
Now (d). If $F^*(G_{\alpha\beta})$ is a $p$-group, then a Sylow $p$-subgroup $P$ of $G_{\alpha\beta}$ is nontrivial. If also $p>2$ and $G_{\alpha\beta}$ is $p$-stable, then
by Glauberman's $Z(J)$-Theorem, $Z(J(P))\triangleleft G_{\alpha\beta}$ for any Sylow $p$-subgroup $P$ of $G_{\alpha\beta}$.  But $Z(J(P))$ is a nontrivial characteristic subgroup of $P$. This contradicts (c).
Now (b). By (a) and solvability, $F^*(G_{\alpha\beta})$ is a $p$-group for some prime $p$. Suppose $p>2$. By (d), $G_{\alpha\beta}$ is not $p$-stable, whence $G_{\alpha\beta}$ has a subquotient isomorphic to $SL_2(p)$.      So $SL_2(p)$ is solvable and $p=3$.
Finally, (e). Suppose that $G_{\alpha\beta}$ has odd order, so it is solvable. By (b) and (d), $F^*(G_{\alpha\beta})$ is a $3$-group and $G_{\alpha\beta}$ is not $3$-stable. Hence $G_{\alpha\beta}$ has an $SL_2(3)$ subquotient, which is of even order, contradiction.
A: An example was constructed by Pablo Spiga:
https://arxiv.org/abs/2102.13614
"A generalization of Sims conjecture for finite primitive groups and two point stabilizers in primitive groups"
