# 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

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What you need is the Third Isomorphism Theorem: given a group $G,$ a normal subgroup $K$ of $G$ and a subgroup $H$ of $G$ we have that $$HK/K \cong H/H\cap K.$$ You rightly guessed that the proof uses the Fundamental Isomorphism Theorem. The homomorphism $f : HK \to H/H \cap K$ defined via $$f(hk) = h (H\cap K)$$ where $h \in H$ and $k \in K$ is a surjective one and its kernel is $K.$
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?
thanks the kernel is $H \cup F_n$. The result then follows – Juan S Apr 7 '11 at 10:25
Sorry, that is $H \cap F_n$! – Juan S Apr 7 '11 at 10:37