Question

Let H and K be subgroups of finite group G such that $K\subseteq H,[G:H]$ is finite and$[H:K]$ is finite. Prove that $[G:K]=[G:H][H:K]$.[Hint: Lagrange]

Short answer

We proved that $[G:K]=[G:H][H:K]$.

Step-by-step solution

To obtain  [G:H]

Let H and K be two subgroups of a finite group G.

Suppose $K\subseteq H$.

Then we have to show that $[G:K]=[G:H][H:K]$.

Using Lagrange’s Theorem, for group G and subgroup H, we get

$\left|G\right|=\left|H\right|[G:H]$

$\Rightarrow [G:H]=\frac{\left|G\right|}{\left|H\right|}$

To obtain [G:K]

From Lagrange’s Theorem, for group G and subgroup K , we get

$\left|G\right|=\left|K\right|[G:K]$

$\Rightarrow [G:K]=\frac{\left|G\right|}{\left|K\right|}$

To obtain [H:K]

From Lagrange’s Theorem, for subgroup H and subgroup K , we get

$\left|H\right|=\left|K\right|[H:K]$

$\Rightarrow [H:K]=\frac{\left|H\right|}{\left|K\right|}$

To show that [G:K ]=[G:H][H:K]

From step1, step 2, and step 3, we can conclude that

$$\begin{gathered} [G:K]=\frac{\left|G\right|}{\left|K\right|} \\ =\frac{\left|G\right|\times \left|H\right|}{\left|H\right|\times \left|K\right|} \\ =\frac{\left|G\right|\times \left|H\right|}{\left|H\right|\times \left|K\right|} \\ =[G:K][H:K] \end{gathered}$$

We proved that$[G:K]=[G:H][H:K]$