Finite index of subgroups If $H1$ and $H2$ are subgroups of finite index of the group $G$ then $$ H1\cap H2$$ is also a subgroup of finite index of $G$. 
How can I prove it?
 A: Index of a subgroup $H < G$ equals exactly cardinality of the set of cosets $G/H$. Therefore it's sufficient to prove, that the set $G/(G_1 \cap G_2)$ is finite. Define map of sets 
$$\Phi \colon G/(G_1 \cap G_2) \to G/G_1 \times G/G_2$$
via formula
$$ \Phi(g \cdot G_1 \cap G_2) = (g \cdot G_1, g \cdot G_2) $$
I leave out the check that $\Phi$ is well-defined.
I claim that $\Phi$ is injective. If $\Phi(g_1 \cdot G_1 \cap G_2) = \Phi(g_2 \cdot G_1 \cap G_2)$, then by definition $g_1 g_2^{-1} \in G_1$ and $g_1 g_2^{-1} \in G_2$, so $g_1 g_2^{-1} \in G_1 \cap G_2$, so $g_1$ and $g_2$ determine the same coset, so we have injectivity.
Now, since we obtained an injective map from set $G/(G_1 \cap G_2)$ into a finite set, we get that $G/(G_1 \cap G_2)$ is finite as well. $\square$ 
A: Consider the action of $G$ on the product $(G/H_1)\times(G/H_2)$.
Here $G/H_1$ isn't the quotient group, but the set of cosets $aH_1$
for $a\in H_1$. The action is defined by $g(aH_1,bH_2)=(gaH_1,gaH_2)$.
This is a finite set, and $H_1\cap H_2$ is the stabiliser of the element
$(H_1,H_2)$.
