In groups, if $H,K$ are subgroups of a finite group $G$, then $[HK:K]=[H:H\cap K]$ where this index counts number of cosets.
In field theory, of $K_1,K_2$ are finite extension of a field $F$ contained in a common field $E$, then it may not be the case that $[K_1K_2:K_1]=[K_2:K_1\cap K_2]$.
Under what conditions on $K_1$ or $K_2$ we can guarantee the equality $[K_1K_2:K_1]=[K_2:K_1\cap K_2]$?
Why this question came? In Fields and Galois theory by Milne, one exercise is to prove: if char$(F)=0$, then $F(x^2)\cap F(x^2-x)=F$, where $F(x)$ is function field in $x$.
But, here we can see that $$F(x^2)F(x^2-x)=F(x), \,\,\,\,\,\,[F(x):F(x^2)]=[F(x):F(x^2-x)]=2$$ and $F(x)$ is Galois over $F(x^2-x)$. Thus, intuitively, if feels that $F(x^2)\cap F(x^2-x)$ is subfield of $F(x)$ with degree $4$ (or less). How this intersection can be $F$, which has infinite index over $F$?