Every subgroup $H$ of a free abelian group $F$ is free abelian I am working through a proof that every subgroup $H$ of a free abelian group $F$ is free abelian (for finite rank)
For the inductive step, let $\{ x_1, \ldots, x_n \}$ be a basis of $F$, let $F_n = \langle x_1,\ldots,x_{n-1} \rangle$, and let $H_n = H \cap F_n$. By induction $H_n$ is free abelian of rank $\le n-1$. Now $$H/H_n = H/(H \cap F_n) \simeq (H+F_n)/F_n \subset F/F_n \simeq \mathbb{Z}$$
The isomorphism I can't see is $$H/(H \cap F_n) \simeq (H+F_n)/F_n.$$ I guess there is a way to get this from the first isomorphism theorem, but I am having a hard time seeing it

Consider the composition
$$
H \stackrel{i}{\hookrightarrow} H + F_n \stackrel{\pi}{\to} (H + F_n)/F_n
$$
where $i$ is the inclusion and $\pi$ the projection. What's its kernel?
