1. ## sub-groups

Is $\displaystyle G$ a finite group , $\displaystyle H$$\displaystyle ,$ $\displaystyle K$ subgroups such that $\displaystyle H K$.
Prove $\displaystyle [ G:H ] = [ G:K ] . [ K:H ]$

Please can develop it in detail.

THANKS YOU FRIENDS....

2. Its just a simple application of Lagrange:

$\displaystyle [G:H]=\frac{ \vert G \vert }{ \vert H \vert }$ and $\displaystyle \vert H \vert = \frac{ \vert K \vert }{ [K:H] }$ substitute in the first and you get $\displaystyle [G:H] = \frac{ \vert G \vert }{ \vert K \vert } [K:H] = [G:K][K:H]$

3. ## help

thanks for the help, that I needed to think a bit