1. ## orders/index, Prove [G:H]=[G:K][K:H]

Suppose H and K are both subgroup of a finite group G, and H⊂K. Prove [G:H]=[G:K][K:H]

2. Originally Posted by apple2009
Suppose H and K are both subgroup of a finite group G, and H⊂K. Prove [G:H]=[G:K][K:H]
In a sense here, we see that $H\leq K\leq G$. It follows by Lagrange's Theorem that both $\left|H\right|$ and $\left|K\right|$ divide $\left|G\right|$.

Therefore,

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

$\left[G:H\right]=\frac{\left|G\right|}{\left|H\right|}\implies\lef t|G\right|=\left[G:H\right]\left|H\right|$

$\left[K:H\right]=\frac{\left|K\right|}{\left|H\right|}\implies\lef t|K\right|=\left[K:H\right]\left|H\right|$.

Substituting the third equation into the first, we get

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

Now substituting this result into the second equation, we have

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

Does this make sense?

3. a less trivial version of this problem is to let G be any group (not necessarily finite) and assume that [G : H] is finite. note that then both [G : K] and [K : H] will also be finite.