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 !
Best Answer
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