I am trying to solve the problem in the title. My attempt is the following:
Let $S$ be a group of order 24 and suppose $G = A_6$ acts transitively on $S$. Then , if we consider some $s \in S$, by the orbit stabilizer theorem, we know that $|\mathcal{O}(s)| = \frac{|G|}{|Stab_G(s)|}$. But then, this implies that $|Stab_G(s)| = 15$ and it is a subgroup of $G$. This is where I'm stuck, since I'm not sure how to continue. I thought that maybe using the fact that $Stab_G(s)$ is cyclic hence abelian (we know this because it is a group of order pq, where p and q are primes and p
Can anyone give me any hint or suggestion on how to continue? Thanks!
Every group of order $15$ is cyclic, and $A_6$ lacks elements of order $15$.
Indeed if $p$ and $q$ are primes with $p<q$ and $q\not\equiv1\pmod p$ then every group of order $pq$ is Abelian, and so cyclic.
To prove this, Sylow's theorems imply that $G$ of order $pq$ has a cyclic normal subgroup of order $q$. Then $G$ will be a semidirect product of this by a cyclic group of order $p$, but that must act trivially on the order $q$ subgroup, as $p\nmid(q-1)$.