An abelian group $G$ is free abelian if and only if satisfies the projective property

Question

I've been reading group theory from Rotman's book "Introduction to the theory of groups" and in Chapter 10 of free abelian groups there is an exercise which am having a hard time to prove.

The exercise is the following:

An Abelian group $F$ is free if and only if it has the projective property.

Necessary definitions:

Free Abelian: An Abelian group $F$ is free abelian if it is the direct sum of infinite cyclic groups.More precisely, there is a subset $X\subset F$ of elements of infinite order called a basis of $F$ with $F=\bigoplus\limits_{x \in X}\langle x\rangle.$

Projective Property: We say that an Abelian group F has the projective property if for every two abelian groups $B,C$ and for every $\beta:B\to C$ surjective homomorphism and every $\alpha:F\to C$ homomorphism there exists homomorphism $\gamma:F\to B$ such that $\alpha = \beta \circ \gamma.$

Rotman has the proof for the $"\implies"$ direction and its understandable although for opposite direction i dont know how to start, i picked $B,C$ to be $\bigoplus\limits_{x \in F}\langle x \rangle$ and $\beta(x)=(\beta_y)_{y \in F}$ where $\beta_y = x$ if $y=x$ and $\beta_y =0$ for $y\neq x$, $\alpha(x) =(\alpha_x)_{x \in F}$ where $\alpha_x =x$ and $\alpha_y = 0 $ for $y\neq x$ and then $\gamma:F \to B$ seems to be an injection hence $F \cong \operatorname{Im}\gamma \leq\bigoplus\limits_{x \in F}\langle x \rangle$ ... is this correct? If not do you have any ideas on how am going to prove it?

Thanks in advance !

Answer 1

Let $A$ be a projective Abelian group, consider the free Abelian group $F_A$ whose basis is $\{[x],x\in A\}$, there exists a canonical projection $p:F_A\rightarrow A$ defined by $p([x])=x$. Since $A$ is projective, there exists $f:A\rightarrow F_A$ such that $p\circ f=Id_A$, we deduce that $f$ is injective and $A$ is isomorphic to a subgroup of $F_A$. A subgroup of a free Abelian group is free.

https://en.wikipedia.org/wiki/Free_abelian_group#Subgroups