Free quotient group admits a section

Statement
Suppose $$N$$ is a normal subgroup of a group $$G$$ such that the quotient group $$G/N$$ is a free group.

Then, $$N$$ is a complemented normal subgroup of $$G$$. In other words, there exists a retract $$B$$ of $$G$$ with normal complement $$N$$, i.e., $$B$$ is a subgroup of $$G$$ such that $$G$$ is the internal semidirect product $$N \rtimes B$$. Explicitly, $$N \cap B$$ is trivial and $$NB = G$$.

A normal subgroup $$N$$ such that $$G/N$$ is a free group is termed a free-quotient subgroup.

Related facts

 * Retract of free group is free on fewer generators