Every subgroup has a left transversal
The proof of this statement requires the axiom of choice. Further information: existence of left transversals is equivalent to axiom of choice
However, for many subgroup, we do not need the axiom of choice. Specifically, what we need is a way to pick representatives for each of the left cosets. We consider the following two cases:
- The subgroup is a subgroup of finite index: In this case, the usual Zermelo-Fraenkel set theory guarantees that we can pick a left transversal. This is because finite choice follows from the ZF axioms.
- The subgroup is a subgroup of countable index: In this case, we only need the axiom of countable choice, which is generally considered weaker and more tenable than the full axiom of choice, since it does not lead to most of the contradictions or paradoxes that arise from the full axiom of choice.
An important related subgroup property is that of being a subgroup having a left transversal that is also a right transversal. Every normal subgroup as well as every subgroup of finite index satisfies this property.