Nonempty intersection of cosets is coset of intersection

DIRECT: The fact or result stated in this article has a trivial/direct/straightforward proof provided we use the correct definitions of the terms involved
View other results with direct proofs
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this

Statement

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: $H_{i}\leq G$, $g_{i}\in G$, $\cap _{i\in I}g_{i}H_{i}$ 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 $\cap _{i\in I}g_{i}H_{i}=u(\cap _{i\in I}H_{i})$.