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Let $F$ be a free abelian group of rank $r$.
Then any subgroup $R$ of $F$ is also a free abelian group of rank at most $r$.

Proof:

Let $F$ be freely generated by $a_1,a_2,\dots,a_r$. We prove by induction on $r$.
If $r=1$ , then $F$ is infinite cyclic.
Then $R$ is either infinite cyclic or $\{0\}$.
Thus $R$ is free Abelian with rank $1$ or $0$.

Suppose the theorem is true for all free Abelian group of rank $<r$.
Let $F$ be a free abelian group of rank $r$.
Define a map $\phi:F\rightarrow \mathbb{Z}$ where $(\mathbb{Z},+)$ is the infinite cyclic group, as follows: $$\phi(k_1a_1+k_2a_2+\dots+k_ra_r)=k_1$$ Then $K=\ker \phi=\{k_2a_2+\dots+k_ra_r\;|\;k_i\in\mathbb{Z},2\leq i \leq r\}$ since $k_1=0$.

Hence $K$ is a subgroup of the free abelian group $\langle a_2,\dots,a_r\rangle$ of rank $r-1$. (*1)
By induction, $K$ is a free abelian group of rank at most $r-1$.

Note that $\phi(R)\subseteq \mathbb{Z}$
Therefore $\phi(R)$ is either infinite cyclic or $\{0\}$.

Case 1: Suppose $\phi(R)=\{0\}$
Then $R\subseteq K$, and by induction, $R$ is a free abelian group of rank at most $r-1$.

Case 2: Suppose $\phi(R)=\langle m \rangle$, $m>0$ Let $r_0\in R$ such that $\phi(r_0)=m$.
We shall prove that $R=\langle r_0\rangle\oplus K$.
Let $r\in R$. Then $\phi(r)=tm$ where $t\in \mathbb{Z}$
Now $r=tr_0+(r-tr_0)$ where $tr_0\in\langle r_0 \rangle$ and $r-tr_0\in K$
So $R=\langle r_0\rangle +K$ (*2)
Let $x\in \langle r_0 \rangle +K$
Then $x=qr_0$. So $\phi(x)=qm$.
But $\phi(x)=0$ since $x\in K=\ker \phi$
This implies that $qm=0$, that is $q=0$, that is $x=0$. Therefore $\langle r_0\rangle \cap K=\{0\}$.
Hence $R=\langle r_0\rangle\oplus K$, a direct sum of an infinite cyclic group and a free abelian group of rank at most $r-1$.
So $R$ is a free abelian group of rank at most $r$.

So far there are two questions.
For (*1), can I straight away say that $K$ is free Abelian group of rank $r-1$? since it itself satisfies the definition without using induction.
For (*2), can we conclude that $R=\langle r_0\rangle\oplus K$? since $K$ is not necessarily a subset of $R$.

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