I am preparing for an algebra prelim at my university, and I came across this question which I cannot finish. Suppose that a finite group $G$ acts transitively on a set $S=\{ a_1,a_2,a_3,a_4,a_5\}$ satisfying the following: $$ Stab(a_1)\cap Stab(a_i) = {e} \textrm{ for all i } \neq 1.$$ Let $H=Stab(a_1)$. Assume that H contains an element of order 4.
(1) Prove that H is cyclic of order 4
(2) Prove that $|G|=20$
Here is my attempt
(1) Let $\psi: G\rightarrow S_5$ be the permutation representation of $G$ on $S$. Then $$Ker(\psi)=\{g\in G|g\cdot a_i = a_i \hspace{3mm} 1\leq i \leq 5\}$$ But $Stab(a_1)\cap Stab(a_i) = {e}$, so this forces $Ker(\psi)=1$. By the first isomorphism theorem and the fact that $Ker(\psi)=\{1\}$, we have $$ G\cong Im(\psi)\leq S_5$$ Now identify $H$ with its isomorphic subgroup of $S_5$ and represent $a_i$ by $i$. $H$ has an element of order 4 that stabilizes $a_1$, so possible generators of $H$ are $\{(2 \hspace{1mm} 3 \hspace{1mm} 4\hspace{1mm} 5),(2\hspace{1mm} 5\hspace{1mm} 4\hspace{1mm} 3), (2\hspace{1mm} 4\hspace{1mm} 3\hspace{1mm} 5), (2\hspace{1mm} 3\hspace{1mm} 5\hspace{1mm} 4),(2\hspace{1mm} 5\hspace{1mm} 3\hspace{1mm} 4), (2\hspace{1mm} 4\hspace{1mm} 5\hspace{1mm} 3) \}$. We could view this set isomorphically as the 6 cycles of length 4 of $S_4$. Now, there are 3 distinct groups that are generated by these 4 cycles. My question is, how can I determine with one $H$ is isomorphic to? $G$ acts transitively on $S$, but I fail to see how this helps since with determining $H$.
(2) Since $G$ acts transitively on $G$, we must have that there is only one orbit. One possibility for $H$ is $\{(2 \hspace{1mm} 3 \hspace{1mm} 4\hspace{1mm} 5), (2\hspace{1mm} 4)(3\hspace{1mm} 5),(2\hspace{1mm} 5\hspace{1mm} 4\hspace{1mm} 3), e \}$. Notice that for any $i$ and $j$ such that $2\leq i, j \leq 5$, there is a element of $H$ that sends $i$ to $j$. So it would seem to me that $H$ acts transitively on $S$. How can I show that $|G|=20$?