Proof of a theorem about cardinality of quotient groups I am looking for a proof of the following proposition: If $H_1<G, H_2<G$ and $[G:H_1]$ and $[G:H_2]$ finite. Then $[G:H_1\cap H_2]$ is also finite
Here's what I've tryed to do:  
$[G:H_1 \cap H_2] = [G:H_1][H_1:H_1 \cap H_2]$
If i show: $[H_1:H_1 \cap H_2] < [G:H_2]$ I can conclude that $[G:H_1 \cap H_2]$ is finite because $[G:H_1][G:H_2]$ is finite.
Let us build the following application: $\phi: \dfrac{H_1}{H_1\cap H_2} \rightarrow \dfrac{G}{H_2}$ such that $x(H_1 \cap H_2) \mapsto xH_2$
**Now all I have to do is show that $\phi$ is injective. How do I do that?
 A: If we take a complete sets of left cosets of $\;H_i,\,i=1,2\;$ in $\;G\;$ , say
$$\left\{\,y_j H_2\,\right\}_{j\in J}\;,\;\;\left\{\,x_i H_1\,\right\}_{i\in I}\;,\;\;|J|,\,\,|I|<\infty$$ 
then for any two indexes $\;i\in I,\,j\in J\;$ we have that either $\;x_iH_1\cap y_jH_2=\emptyset\;$ or else the $\;x_iH_1\cap y_jH_2\;$ is a coset in $\;H_1\cap H_2\;$ , because:
$$z\in x_1H_1\cap y_jH_2\implies\begin{cases}zH_1=x_iH_1\\{}\\zH_2=y_jH_2\end{cases}\;\implies x_iH_1\cap y_jH_2=zH_1\cap zH_2$$
and we get that
$$t\in x_iH_1\cap y_jH_2=zH_1\cap zH_2\iff t=zh_1=zh_2\;,\;\;h_i\in H_i \iff$$
$$\iff z^{-1}t=h_1=h_2\in H_1\cap H_2\iff t= z(H_1\cap H_2)$$
Try now to round up the last details of the proof.
A: Let $H = H_1 \cap H_2$.
Let $gH$ by any (left) coset of $H$ ($g \in G$).
Claim: $gH = (gH_1) \cap (gH_2)$.
Proof: Every element of $gH$ is of the form $gh$, for some element $h$ that belongs to both $H_1$ and $H_2$, which proves that every $gH \subseteq (gH_1) \cap (gH_2)$. On the other hand, any element of $(gH_1) \cap (gH_2)$ is of the form $gh_1$, $\exists h_1 \in H_1$, as well as $gh_2$, $\exists h_2 \in H_2$, but then $gh_1 = gh_2 \implies h_1 = h_2 \in H_1 \cap H_2 \implies gh_1 = gh_2 \in gH$. Thus, $(gH_1) \cap (gH_2) \subseteq gH$, and the claim is proved.
Since every left coset of $H$ is the intersection of some left coset of $H_1$ and some left coset of $H_2$, and the number of left cosets of $H_1$ and $H_2$ are both finite, it follows that the number of left cosets of $H$, i.e., $[G : H_1 \cap H_2]$ is also finite.
A: You could use a general result, that tells you that, given subgroups $H_{1}, H_{2}$ of the group $G$, there is a bijection between
$$
\frac{H_{1}}{H_{1} \cap H_{2}} \qquad\text{and}\qquad \frac{H_1 H_{2}}{H_{2}},
$$
where the second fraction denotes the set of cosets $g H_{2}$, where $g$ lies in the set $H_{1} H_{2} = \{ x y : x \in H_{1}, y \in H_{2} \}$.
To prove this consider the map
$$
f : H_{1} \to \frac{H_{1} H_{2}}{H_{2}}
$$
given by $x \mapsto x H_{2}$, which is clearly surjective. We have $f(x) = f(x')$ iff $x H_{2} = x' H_{2}$ iff $x^{-1} x' \in H_{2}$ iff $x^{-1} x' \in H_{1} \cap H_{2}$ iff $x (H_{1} \cap H_{2}) = x' (H_{1} \cap H_{2})$.
Thus $f$ induces a bijection
$$
\begin{cases}
\dfrac{H_{1}}{H_{1} \cap H_{2}} \to \dfrac{H_1 H_{2}}{H_{2}}\\
x (H_{1} \cap H_{2}) \mapsto x H_{2}
\end{cases}
$$
