We can think $S_5$ acting on the field $F_5$ with five elements. I define
$C_5 = \{ \sigma : \sigma(x) = x + b; b \in F_5 \}$
The problem is to show that $S_5$, $A_5$, $N_5$, $D_5$ and $C_5$ are the only subgroups containing $C_5$, where $N_5$ is the normalizer of $C_5$ in $S_5$ and $D_5$ is the dihedral group with $\sigma$ s.t., $\sigma(x) = \pm x + b; b \in F_5$.
I have many problems to do it. To start, I proved if $H$ is a subgroup cotaining $C_5$, its order can be 5, 10, 15, 20, 30, 40, 60 and 120. But, in $S_5$ there is no subgroups with some of these orders.
Thanks for your help!
Answering this is easier if we use the fact that $A_5$ is a simple group, from which it follows easily that the only normal subgroups of $S_5$ are $1$, $A_5$, and $S_5$.
So $S_5$ cannot have subgroups of index $3$ or $4$, since such subgroups would imply homomorphisms onto transitive subgroups of $S_3$ and $S_4$. Also groups of order $15$ are cyclic, and $S_5$ has no element of order $15$, so it has no such subgroups.
This leaves orders $5$, $10$, $20$, $60$ and $120$ as possibilities. $S_5$ and $A_5$ are the only subgroups of orders $120$ and $60$. Groups having orders $10$, $20$ or $40$ have normal Sylow $5$-subgroups, so they are contained in the normalizer in $S_5$ of the subgroup $\langle (1,2,3,4,5) \rangle$, which is the subgroup $\langle (1,2,3,4,5), (2,3,5,4) \rangle$ of order $20$.