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 $u v$ is in $g_{i} H_{i}$ . Hence $u v$ 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 $\cap_{i \in I} g_{i} H_{i} = u (\cap_{i \in I} H_{i})$ .