Let $G$ be an abelian group and $H\le G$. Then TFAE:
- $G/H$ is free.
- $H$ is a direct summand of $G$.
This is how I tried to prove this theorem:
If $G/H$ is free, then $G/H$ has a basis. Let's denote this basis as $(g_1H, g_2H, ..., g_nH)$, where each $g_i$ is an element of $G$.
We need to find a subgroup $K$ such that : $G=H\oplus K$. Let $K=\langle g_1,g_2\dots,g_n\rangle $
First, we need to show that $H\cap K=\{0\}$.
Let $h\in H\cap K$, then $h\in H$ and $h=\sum k_ig_i$ where $k_i\in \mathbb{Z}$.
$h\in H\Rightarrow \bar{h}=hH=H$, which implies that $h$ is congruent to the identity element modulo $H$.
Then
$$\begin{align} \bar{h}=\bar{\sum k_ig_i}=\bar{0} &\Rightarrow k_1g_1H+\dots +k_ng_nH=0\\ &\Rightarrow k_1=k_2=\dots =k_n=0\\ &\Rightarrow h=0, \end{align}$$
then $H\cap K=\{0\}$.
Next we need to show that every element $x$ of $G$ can be written as $x=h+k$ where $h\in H$ and $k\in K$.
Let $x\in G$, then $\bar{x}=x+H\in G/H\Rightarrow x+H=k_1(g_1+H)+\dots +k_n(g_n+H)$ where $k_i\in \mathbb{Z}$, then $x+H=(k_1g_1+\dots k_ng_n)+H\Rightarrow (k_1g_1+\dots k_ng_n)-x\in H$
$\exists h\in H$ such that $k_1g_1+\dots k_ng_n-x=h\Rightarrow x=(-h)+(k_1g_1+\dots +k_ng_n)\in H+K$
Seconde implication :
$H$ is a direct summand of $G$, there existe $K\leq G$ such that $G=H\oplus K$. Let $xH\in G/H$, where $x\in G$, then $\exists !h\in H$ and $k\in K$ such that $x=h+k$.
$xH=x+H=(h+k)+H=k+(h+H)k+H\Rightarrow \bar{x}=\bar{k}$.
Let {$k_iH$}$_i$ be all the distincts cosets generated by $K$, then $G/H=<k_iH>$.
Next we need to show that this set is linear independence.
If $\sum_i d_i(k_i+H)=\bar{0}=H$ where $d_i\in \mathbb{Z}$, then $$\sum_i d_i(k_i+H)=\sum_i(d_ik_i)+H=H \Rightarrow \sum_i d_ik_i\in H\Rightarrow \sum_i d_ik_i=0$$
But how can I show that $d_i=0,\forall i$