 Solution

Let $\pi :G\to G/K$ be the quotient group homomorphism. Its restriction to $X$ is injective because its kernel there is $K\cap X=\{1\}$. Since $g\in K$, $\pi (x_{i}g)=\pi (x_i)$ for every generator. Let $F_n$ be the abstract free group on letters $a_1,\dots ,a_n$ and define the surjective group homomorphism $\alpha :F_n\to ⟨x_{1}g,\dots ,x_{n}g⟩$ by $\alpha (a_i)=x_{i}g$. If $w\in \ker \alpha$, thenInjectivity on $X$ implies $w(x_1,\dots ,x_n)=1$ in $X$. Those elements are a free basis of a group, so $w$ is the empty reduced word in $F_n$. Thus $\ker \alpha =\{1\}$. This is freeness detected by a quotient, and proves