prove $[E:K\cap L]=[E:K][E:L]$ Let $E/K,E/L$ be finite Galois extensions with Galois groups $G,H$. If $G\cap H= \left\{1\right\}$, show $G\cdot H = \left\{ g h \mid g\in G,h\in H \right\}$ is a group iff $[E:K\cap L]=[E:K][E:L]$.
But isn't the set $G\cdot H$ always a group? The condition $G\cap H= \left\{1\right\}$ implies $KL=E$ and then there's a theorem saying $[KL:L]=[K:L\cap K]$ which is different...
 A: There is something wrong with the title of this question. Namely the claim in the title is false without some extra assumptions.
Consider the following example.
Let $E=\Bbb{C}(x)$ be the field of rational functions. Let $G$ be the cyclic group of order two generated by the $\Bbb{C}$-automorphism $\sigma:x\mapsto -x$. We immediately see that the fixed field of $G$ is the subfield $K=\Bbb{C}(x^2)$.
Let $H$ be the cyclic group of order two generated by the $\Bbb{C}$-automorphism $\tau:x\mapsto 1-x$. We easily see that the corresponding fixed field is $L=\Bbb{C}(x^2-x)$.
But in this case it is easy to show that
$$
K\cap L=\Bbb{C}.
$$
See this thread and my old answer for a proof of that fact. Anyway, here
$$
[E:L]=[E:K]=2,\quad [E:K\cap L]=\infty.
$$
Note that in this case $\sigma\tau:x\mapsto x+1$, so $\sigma\tau$ is of infinite order. Therefore $GH$, a set with only four elements, most certainly is not a group.

Answering one direction. Assume that $G\cdot H$ is a group, call it $\Omega$. Because $G\cap H=\{1\}$ we know that $|\Omega|=|G|\cdot |H|$, in particular we know that $\Omega$ is finite. By Artin's lemma the fixed field $F=E^\Omega$ has the properties that
$[E:F]=|\Omega|$ and $Gal(E/F)=\Omega$. Because $\Omega$ is the smallest group containing both $G$ and $H$, Galois correspondence tells us that $F$ is the largest subfiel contained in both $K$ and $L$. Therefore $F=K\cap L$. Also,
$$
[E:K\cap L]=[E:F]=|\Omega|=|G|\cdot |H|=[E:K]\cdot [E:L].
$$

[Edit:]
I think the other direction follows from the following line of thought. Assume that $[E:K\cap L]=|G|\cdot |H|$. Clearly $K\cap L$ is fixed pointwise by all the automorphisms of $E$ that are of the form $gh$. Because $G$ and $H$ intersect trivially, all the automorphisms of this form are distinct. Therefore
$|Aut(E/K\cap L)|\ge |G|\cdot |H|$. On the other hand for a finite extension $E/F$ we have $|Aut(E/F)|\le [E:F]$ and here there is equality if and only if $E/F$ is Galois. Therefore we can conclude that $E/(K\cap L)$ is Galois, and
$Aut(E/K\cap L)=G\cdot H$. In particular, $G\cdot H$ is a group.
