I'm proving a result due to Neumann (1954) and require a certain step which I am not able to get. I am certain that it is true, since it appears in an outline suggested by our teacher.
Let $L_1,...,L_n$ be subgroups of $G$ of finite indexes and $K = \bigcap_{i=1}^n L_i$. Prove that there exists $y_1,...,y_m \in G$ such that $$\bigcup_{i=1}^n x_iL_i \subseteq \bigcup_{j=1}^m y_j K$$
I have tried doing a giant reunion by writing each $L_i$ as $L_i\cap G$ and then replacing $G$ by a reunion of left classes of other $L_j$. The issue is that this proof would lead me to saying that the entirety of $G$ can be written as the reunion of left cosets of $K$, which seems too strong of a result.
I have also proven that if I do get the result for $n=2$, then I have it for all $n$ by induction. However, even this case I can't find...
Finally, I have also followed the reasoning: $$\bigcup_{j=1}^m y_j K = \bigcap_{i=1}^n \bigcup_{j=1}^m y_j L_i$$ But this doesn't really suggest an expression for the $y_j$ factors...
As pointed out, we simply have to prove that $K$ is of finite index in each $L_i$. Also, it only has to be proven in the case $n=2$, other cases being immediate through induction.
It can be written $G=a_1L_1\cup...\cup a_pL_1$ and therefore:
$$L_2=(a_1L_1\cap L_2)\cup...\cup(a_pL_1\cap L_2)$$
Eliminating empty sets, one can pick an element $b_i\in a_iL_1\cap L_2$ and therefore write:
$$L_2=b_1(L_1\cap L_2)\cup...\cup b_r(L_1\cap L_2)$$
Therefore proving $K=L_1\cap L_2$ is of finite index in $L_2$ (and therefore in $L_1$).