Let $p\neq5$ be a prime number such that $q=(p-1)/2$ is prime. Does there exist an Alternating group of degree $p$ in which every minimal subgroup of order $p$ is contained properly in exactly one maximal subgroup of $A_{p}$?

Note that if such Alternating groups exist, then the mentioned maximal subgroup is isomorphic to $\mathbb{Z}_{p}\rtimes\mathbb{Z}_{q}$


Yes, this is the case for most primes of this form. The smallest example is with $p=47$.

Burnside proved that a transitive group of prime degree $p$ either has a normal subgroup of order $p$ or is doubly transitive. As a consequence of the classification of finite simple groups, the doubly transitive groups are now known.

The ones with prime degree other than $A_p$ and $S_p$ are:

(i) ${\rm PSL}(2,11)$ and $M_{11}$ with $p=11$;

(ii) $M_{23}$ with $p=23$.

(iii) Groups $G$ with ${\rm PSL}(d,q) \le G \le {\rm P \Gamma L}(d,q)$, where $p = (q^d-1)/(q-1)$ is prime.

So for all primes not of this form, the only maximal subgroup of $A_p$ containing a subgroup $P$ of order $p$ is $N_G(P) = C_p \rtimes C_q$.

This is the case for $q=43,59,83,107,\ldots$.

In fact in Case (iii), $(p-1)/2 =(q^d-q)/(2(q-1))$ is prime only when $p=7$.


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