Why is $H_1 \le G \land H_2 \le G$ necessary in $a(H_1 \cap H_2) = aH_1 \cap aH_2$? The problem is as follows:

$G$ is a group and $H_1$ and $H_2$ are its two subgroups (i.e., $H_1 \le G \land H_2 \le G$). To prove that $a(H_1 \cap H_2) = aH_1 \cap aH_2$.

Here is my trial:
On the one hand, for $a(H_1 \cap H_2) \subseteq aH_1 \cap aH_2$:
$b \in a(H_1 \cap H_2) \implies \exists h \in H_1 \cap H_2: b = ah \implies b \in aH_1 (h \in H_1) \land b \in aH_2 (h \in H_2) \implies b \in aH_1 \cap aH_2.$
On the other hand, for $aH_1 \cap aH_2 \subseteq a(H_1 \cap H_2)$:
$b \in aH_1 \cap aH_2 \implies b \in aH_1 \land b \in aH_2 \implies \exists h_1 \in H_1, h_2 \in H_2: b = ah_1 \land b = ah_2 \implies h_1 = h_2 \implies \exists h (= h_1 = h_2) \in H_1 \cap H_2: b = ah \implies b \in a(H_1 \cap H_2).$
In this argument, I do not (in my opinion) use the assumption of $H_1 \le G \land H_2 \le G$. Therefore, 

What is wrong with my argument?
  Why is $H_1 \le G \land H_2 \le G$ necessary in $a(H_1 \cap H_2) = aH_1 \cap aH_2$?

 A: The need for the $H_1 \le G \land H_2 \le G$ condition hinges on the definition used for expressions of the form $aH$.  In all cases, $aH$ is defined as
$$aH := \{ah\;|\;h \in H\}$$
but some textbooks may limit this definition only to those $H$ that are also subgroups.  I imagine this is the case whenever expressions like $aH$ are used as shorthand for "left coset", which would make sense only if $H$ is a subgroup.
In fact, a proof having the same structure as yours shows that, for any injective function $f$, and any subsets $A$ and $B$ of $\mathrm{dom}(f)$,
$$f(A \cap B) = f(A) \cap f(B).$$
(Actually, $f(A \cap B) \subseteq f(A) \cap f(B)$ holds for any function $f$, but the reverse inclusion holds only if $f$ is injective.)
Now, for any $a \in G$, define $f_a: x \mapsto ax$, and your statement follows as a corollary of the equality above.  (Every such $f_a$ is injective, and in fact bijective, since $f_a^{-1} = f_{a^{-1}}$.  BTW, this injectivity condition makes its appearance in your proof in the step $b = ah_1 \land b = ah_2 \implies h_1 = h_2$.)
