Prove that any subgroup of a free Abelian group of rank $r$ is free Abelian of rank at most $r$. 
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 $x\in \langle r_0 \rangle \cap K$
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 $R$.
 A: *

*Actually, $K=\langle a_2,\ldots a_r\rangle$; this is a subset of the free generators for $F$, so the conclusion is that $K$ is free of rank $r-1$:  the induction hypothesis doesn't come into it. In general, if $F$ is free on $S$, and $T\subseteq S$, then $\langle T\rangle$ is free on $T$ (this is very easy to prove using for example the universal property, but it is also easy using the explicit description as the direct sum of copies of $\mathbb{Z}$ or any other explicit construction you have).


*To prove that $C=A\oplus B$, you need to prove two things: you need to prove that $A\cap B=\{0\}$, and that $C=A+B$. Your argument only contains a proof that $\langle r_0\rangle\cap K=\{0\}$. Even on its face, it is incomplete to establish your claimed conclusion. And in fact, you are correct to be worried: normally, in proving $C=A\oplus B$, we start with $A$ and $B$ subgroups of $C$, so that the inclusion $A+B\subseteq C$ is never mentioned (because it is obvious), and so we only need to check that $C\subseteq A+B$. Here, you do not have that $K\leq R$ (most of the time it won't be), so you cannot show that $R=\langle r_0\rangle + K$, which is the step you are missing.
What you need to prove is that $R=\langle r_0\rangle \oplus (K\cap R)$. That the intersection is trivial follows by what you've already done (or by the exact same argument). To prove that $R=\langle r_0\rangle +  (K\cap R)$, let $r\in R$. Then find a multiple of $r_0$ such that $r-mr_0\in K$ and go from there to finish it off.
