Let $P$ be a non-abelian finite $p$-group which is a transitive permutation group of degree $p^n$ such that the stabiliser of a point is meet-irreducible.
Suppose $P$ has a non-transitive abelian maximal subgroup $M$.
I am told that if $S$ is a stabiliser of a point in $P$ then $S\le M$, since $M$ is non-transitive and I am trying to understand why?
Obviously if $S\not\le M$ then $<S,M>=P$ which is transitive. Is there a contradiction here?
Answer provided by Derek Holt in comments (rewritten and details added here to get full answer):
Maximal subgroups of $p$-groups have index $p$. Suppose $S=Stab_G(x)$ and notice $Stab_M(x)=S\cap M$, then by the Orbit-stabilizer Theorem $|G|=|S||x^G|$ and $|M|=|M\cap S||x^M|$. This gives $p=|G:M|=\frac{|G|}{|M|}=\frac{|S||x^G|}{|M\cap S||x^M|}$. Since $G$ is transitive and $M$ is intransitive $\frac{|x^G|}{|x^M|}> 1$, which is only possible if $|S|=|S\cap M|$. Hence $S\le M$.