Finite Index of Subgroup of Subgroup Prove the following: If $H$ is a subgroup of finite index in a group $G$, and $K$ is a subgroup of $G$ containing $H$, then $K$ is of finite index in $G$ and $[G:H] = [G:K][K:H]$.
So this is basically a bijective proof? The number of cosets of $H$ in $G$ equals the number of cosets of $K$ in $G$ times the number of cosets of $H$ in $K$ by the multiplication principle? 
Reference: Fraleigh p. 103  Question 10.35 in A First Course in Abstract Algebra

While studying today, I think I did a very similar exercise. It is presented as:

Being $G$ a finite group and $H \leq K \leq G$, prove that $|G:H| = |G:K|\cdot|K:H|$.

I answered the following:

$|G:K|\cdot|K:H| = \frac{|G|}{|K|}\cdot\frac{|K|}{|H|} = \frac{|G|\cdot|K|}{|K|\cdot|H|} = |G:H|$

Is it the same thing or am I mistaking anything?
Thank you for the attention.
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