$H_1,H_2$ subgroups such that $|H_2|/|H_1|$ is a prime. There is no subgroups stricly between them.

Question

How could one begin by demonstrating this? I really can't think of a strategy to leave.

Let $G$ a finite group. Let $H_1$, $H_2$ subgroups of $G$ such that $H_1\subset H_2$ and $|H_2|/|H_1|$ is prime. Show that there is no subgroup stricly between $H_1$ and $H_2$.

Answer 1

Let $H$ be subgroup between subgroups $H_1$ and $H_2$.So, $H_1 \leq H \leq H_2$. Then, $[H_2:H][H:H_1]=[H_2:H_1]=|H_2|/|H_1|$ which is prime. So $[H_2:H]=1$ or $[H_1:H]=1$ So $H_2=H$ or $H_1=H$

Answer 2

(Big) hint. Let $K$ be a subgroup such that $H_1\subset K\subset H_2$. We know that $[H_2:H_1]=p$ for $p$ prime. What can the indices

$$[H_2:K]\quad \text{and} \quad [K:H_1]$$

now be? (Use the tower law for subgroups.)