Basis for a free abelian group $F(X)/F(X)'$

Question

Let $|X|=n,$ prove that $F(X)/F(x)'$ is a free abelian group of rank $|X|$. For the rank prove and use that $\{xF(X)' : x \in X \}$ is a basis for $F(X)/F(X)'$.

The abelian part is because in a group $G$, $G'\lhd G $, and $G/G'$ is abelian.

I try to prove the basis and I get a lost, because I don't know how to proced from this:

Let $k_i \in \mathbb{Z}$, then if $k_1(x_1F(X)')+ \cdots k_n(x_nF(X)')=0=F(X)'$.

Answer 1

If $A$ is an abelian group (let me write it with additive notation), and $B \subseteq A$, then $B$ is a basis for $A$ if the following holds:

For any abelian group $G$, and any function $f : B \to G$, there exists a unique group homomorphism $\overline f : A \to G$ such that $\overline f|_B = f$.

Proof. First, we will prove that $B$ generates $A$:

• If $\langle B \rangle$ is the generated subgroup of $A$ by $B$, let $$B \stackrel i\hookrightarrow \langle B \rangle \stackrel j\hookrightarrow A$$ be the natural inclusions. By assumption, there is a group homomorphism $\overline i : A \to \langle B \rangle$ such that $\overline i|_B = i$, but then $(j \circ \overline i)|_B = j \circ \overline i|_B = j \circ i$, and so $j \circ \overline i = \operatorname{id}_A$ (the identity map is clearly the unique group homomorphism $\overline{j \circ i} : A \to A$ such that $(\overline{j \circ i})|_B = j \circ i$). Thus $j$ is surjective, and $A = \langle B \rangle$.

Now, $B$ is linearly independent:

• Let $k \in \mathbb Z^+$ and $b_1,\dots,b_k \in B$ such that $n_1b_1 + \cdots + n_kb_k = 0$ for some $n_1,\dots,n_k \in \mathbb Z$. Consider the abelian group $\mathbb Z^{\oplus B}$ of all functions $B \to \mathbb Z$ with finite support, and let $\chi : B \to \mathbb Z^{\oplus B}$ be the function that sends $b \in B$ to the characteristic function $\chi_b \in \mathbb Z^{\oplus B}$, i.e., the function $\chi_b : B \to \mathbb Z$ such that $$(\forall x \in B) \quad \chi_b(x) = \begin{cases} 1 & \textrm{if } x = b, \\ 0 & \textrm{if } x \neq b. \end{cases}$$ Now, by assumption, there is a group homomorphism $\overline \chi : A \to \mathbb Z^{\oplus B}$ so that $\overline \chi|_B = \chi$, and so if we apply $\overline \chi$ on both sides of $n_1b_1 + \cdots + n_kb_k = 0$ we get that $n_1\chi_{b_1} + \cdots + n_k \chi_{b_k} = 0$. It follows that $$(\forall j \in \{1,\dots,k\}) \quad 0 = 0(b_j) = (n_1\chi_{b_1} + \cdots + n_k \chi_{b_k})(b_j) = n_j.$$

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Once we observed this, for $A = F(X)/F(X)'$ and $B = \{xF(X)' : x \in X\}$ prove that the quoted property above holds (as a hint, use the universal property of the free group on $X$).

Answer 2

No calculation is necessary and a much more conceptual proof is possible using some ingredients from category theory:

1. The functor $(-)^{ab} : \mathbf{Grp} \to \mathbf{Ab}$, $G \mapsto G/G'$ is left adjoint to the forgetful functor. This is the usual universal property of the abelianizaton of a group.

2. The functor $F : \mathbf{Set} \to \mathbf{Grp}$, $X \mapsto F(X)$ is left adjoint to the forgetful functor. This is the usual universal property of the free group.

3. The functor $F^{ab} : \mathbf{Set} \to \mathbf{Ab}$, $X \mapsto \mathbb{Z}^{\oplus X}$ is left adjoint to the forgetful functor. This is the usual universal property of the free abelian group.

4. The following general fact: Adjoint functors compose: If $F$ is left adjoint to $G$ and $F'$ is left adjoint to $G'$, then $F \circ F'$ is left adjoint to $G' \circ G$ (assuming the target of $F'$ is the source of $F$, so that $F \circ F'$ is defined).

5. The following general fact: Adjoint functors are unique up to isomorphism. This is a consequence of the Yoneda Lemma.

It follows that the composition $(-)^{ab} \circ F$ is left adjoint to the forgetful functor $\mathbf{Ab} \to \mathbf{Set}$, and since $F^{ab}$ is also left adjoint to that forgetful functor, they must be isomorphic. Concretely, it means that the canonical morphism $F^{ab}(X) \to F(X)^{ab}$ is an isomorphism, so that $F(X)^{ab}$ is free abelian.

What is important to realize is that 1-5. are not just statements which are used in the proof here, but are used in lots of other situations as well, in particular 4. and 5. For example, it immediately implies that the monoid ring of a free commutative monoid is a polynomial ring (=free commutative ring). Category theory allows us to see this directly without any extra calculation, and we have also captured the general pattern behind these isomorphisms.