Nonempty intersection of cosets is coset of intersection

Verbal statement
If the intersection of a collection of left cosets of subgroups is nonempty, then it is a coset of the intersection of the corresponding subgroups.

Statement with symbols
Suppose $$\{H_{i}\}_{i \in I}$$ is a family of subgroups of a group $$G$$ indexed by $$I$$, and $$g_{i}$$ are elements of $$G$$. Then $$\cap_{i \in I}g_{i}H_{i}$$, if non-empty, is a left coset of the subgroup $$\cap_{i \in I}H_{i}$$.

Related facts

 * Subgroup containment implies coset containment: If one subgroup of a group is contained in another, then every left coset of the subgroup is contained in a left coset of the other subgroup.
 * Coset containment implies subgroup containment: If a left coset of one subgroup is contained in a left coset of another, then the subgroup containment also holds.

Proof
Given: $$H_{i} \le G$$, $$g_{i}\in G$$, $$\cap_{i \in I}g_{i}H_{i}$$ non-empty.

To prove: there exists $$a \in G$$ such that $$\cap_{i \in I}g_{i}H_{i}=a\cap_{i \in I}H_{i}$$

Proof: Observe that for any $$u$$ in $$\cap_{i \in I}g_{i}H_{i}$$, we have $$g_{i}^{-1}u \in H_{i}$$, viz: $$u^{-1}g_{i} \in H_{i}$$. So, $$u^{-1}g_{i}H_{i}=H_{i}$$.

For any $$v \in \cap_{i \in I}H_{i}$$, in each $$H_{i}$$ we can find $$h_{i}$$ such that $$v=u^{-1}g_{i}h_{i}$$. Therefore $$uv$$ is in $$g_{i}H_{i}$$. Hence $$uv$$ is in $$\cap_{i \in I} g_{i}H_{i}$$, viz: $$u(\cap_{i \in I}H_{i}) \subseteq \cap_{i \in I} g_{i}H_{i}$$.

Now, for all $$u, p$$ in $$\cap_{i \in I}g_{i}H_{i}$$, we can find $$h_{i}, k_{i} \in H_{i}$$ such that $$u=g_{i}h_{i}$$ and $$p=g_{i}k_{i}$$. Then $$p^{-1}u=k_{i}^{-1}g_{i}^{-1}g_{i}h_{i}=k_{i}^{-1}h_{i} \in H_{i}$$. So it follows that the cosets of the intersection subgroup with respect to $$u, p$$ are the same. Therefore, $$\cap_{i \in I} g_{i}H_{i} \subseteq u(\cap_{i \in I}H_{i})$$. Hence $$\cap_{i \in I} g_{i}H_{i} = u(\cap_{i \in I}H_{i})$$.