Question

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

Short answer

We proved that, $\left[G:K\right]=\left[G:H\right]\left[H:K\right]$

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 $\left[G:K\right]=\left[G:H\right]\left[H:K\right]$.

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

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

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

To obtain  [G : K]

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

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

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

To obtain [H : K]

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

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

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

To show that [G : K]=[G : H][H : K]

From step 1, step 2 and step 3, we can conclude that,

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

We proved that, $\left[G:K\right]=\left[G:H\right]\left[H:K\right]$