First I must point out the following corollary in my book:
If $G$ is a finitely generated abelian group generated by $n$ elements, then every subgroup $H$ of $G$ may be generated by $m$ elements with $m \le n$.
Here is the problem I am working on:
If $F$ is a finitely generated free abelian group generated by $n$ elements, $F$ has rank $m \le n$.
My question is, is the following are argument valid?
Since $F$ is a subgroup of itself, and by hypothesis $F$ can be generated by $n$ elements, it follows that $F$ as a subgroup can be generated Since any basis $X$ of $F$ by definition generates $F$, it follows that $X$ can contain no more than $m$ elements and therefore $rank ~(F) = |X| \le m \le n$.
I ask about its validity, because I found this (first problem on the first page) seemingly unnecessarily complicated argument, and it has cast doubt on my own solution. Here is it:
If $F$ is finitely generated by $n$ elements $x_1,...,x_n$. Let $X = \{x_1,...x_n \}$; let $i : X \to F$ denote the inclusion map; and let $F(X)$ denote the free abelian group on $X$ with morphism $\iota X \to F(X)$. Then by the property of free objects, there exists an abelian group homomorphism $p : F(X) \to F$ such that $p \circ \iota = i$. The assumption that $F$ is generated by $X$ implies that $p$ is an epimorphism. Then $F(X)/ \ker p \simeq F$. If $F$ has rank $m$, then it has a basis $\{y_1,...,y_m\}$. Thus $F(X) = \ker p \oplus \langle p^{-1}(y_1),...,p^{-1}(y_m) \rangle$. The subgroup $\ker p$ of $F(X)$is also free abelian. Suppose that $\ker p$ has rank $r$. Then $n = r + m$, so $m \le n$.
So, is this proof unnecessarily complicated; is mine correct? Also, does this solution presuppose that a subgroup of a free group is free, or is there a way specifically to show that $\ker p$ is free? I ask because I don't believe it has yet been proven in my book that a subgroup of a free group is free.
EDIT:
So it appears that the solution presupposes that a subgroup of free abelian group is free abelian. Is there any proof that avoids this fact? Regarding the solution, why is it that $p : F(X) \to F$, the unique group homomorphism guaranteed to exist, is surjective? The solution says that it is due to the fact that $X$ generates $F$, but isn't this assuming something like $p(x_i) = x_i$ is true, which would allow us to say $p(\langle X \rangle) = \langle p(x_1),...,p(x_n) \rangle = \langle x_1,...,x_n \rangle = F$? I don't see why $p(x_i) = x_i$ is true, however.
Let $f : F(X) \to F$ be the homomorphic extension of $F(X) \ni x_i \mapsto x_i \in F$. In order to show that $p(x_i) = x_i$, it seems that I would need to show that $f$ satisfies $f \circ \iota = i$, as uniqueness would force $p=f$, but I don't see how to prove that $f \circ \iota = i$, particularly since I don't know how $\iota$ maps elements.