Nonempty intersection of cosets is coset of intersection

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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


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}).