# Subgroup of finite index has a left transversal that is also a right transversal

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., subgroup of finite index) must also satisfy the second subgroup property (i.e., subgroup having a left transversal that is also a right transversal)
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## Statement

Suppose $G$ is a group and $H$ is a subgroup of finite index in $G$, i.e., the index of $H$ in $G$ is finite. Then, there exists a subset $S$ of $G$ such that $S$ is both a left transversal and a Right transversal (?) of $H$ in $G$. In other words, $S$ intersects every left coset of $H$ in exactly one element, and it also intersects every right coset of $H$ in exactly one element.

## Related facts

### Other conditions that guarantee the existence of a left transversal that is also a right transversal

Property Meaning Proof of implication
normal subgroup each left coset is a right coset
permutably complemented subgroup there is a permutable complement
subgroup of finite group the whole group is finite subgroup of finite group has a left transversal that is also a right transversal
subset-conjugacy-closed subgroup subset-conjugacy-closed implies left transversal that is also a right transversal
retract has a normal complement (via subset-conjugacy-closed, via permutably complemented)

## Proof

### Proof outline

We use fact (2) to quotient out by the normal core, and we use fact (1) to prove the result in the quotient and then pull back to the original group.