Suppose H and K are both subgroup of a finite group G, and H⊂K. Prove [G:H]=[G:K][K:H]
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In a sense here, we see that $\displaystyle H\leq K\leq G$. It follows by Lagrange's Theorem that both $\displaystyle \left|H\right|$ and $\displaystyle \left|K\right|$ divide $\displaystyle \left|G\right|$.
Therefore,
$\displaystyle \left[G:K\right]=\frac{\left|G\right|}{\left|K\right|}\implies \left|G\right|=\left[G:K\right]\left|K\right|$
$\displaystyle \left[G:H\right]=\frac{\left|G\right|}{\left|H\right|}\implies\lef t|G\right|=\left[G:H\right]\left|H\right|$
$\displaystyle \left[K:H\right]=\frac{\left|K\right|}{\left|H\right|}\implies\lef t|K\right|=\left[K:H\right]\left|H\right|$.
Substituting the third equation into the first, we get
$\displaystyle \left|G\right|=\left[G:K\right]\left[K:H\right]\left|H\right|$.
Now substituting this result into the second equation, we have
$\displaystyle \left[G:K\right]\left[K:H\right]\left|H\right|=\left[G:H\right]\left|H\right|\implies \left[G:H\right]=\left[G:K\right]\left[K:H\right]$.
Does this make sense?