# Nonempty intersection of cosets is coset of intersection

## Contents

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## 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 $\cap_{i \in I}H_{i}$.

## 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 $\cap_{i \in I} g_{i}H_{i} = u(\cap_{i \in I}H_{i})$.