Showing that $\langle A \rangle$ can be equivalently defined via $F(A)$ and via $F^{ab}(A)$ when $G$ is abelian, $A \subset G$.

Question

In Aluffi's Algebra: Chapter 0 the author introduces the subgroup generated by a subset as follows (paraphrased):

Let $G$ be a group, and $A \subset G$ any subset. By the universal property of free groups there is a unique group homomorphism $$\varphi: F(A) \to G$$ extending this inclusion. The image of this homomorphism is a subgroup of $G$, the subgroup generated by $A$ in $G$, often denoted $\langle A \rangle$.

He then states the following:

Of course, if $G$ is abelian, then $\varphi$ factors through $F^{ab}(A)$, so we may replace $F(A)$ by $F^{ab}(A)$ in this case.

While this seems plausible enough, I would like to verify it rigorously, and this is where I am a bit stuck at the moment. I would prefer to do it without invoking any concrete construction of free groups. The following is my attempt so far.

Let $j: A \to F(A)$ and $k: A \to F^{ab}(A)$ be the set-functions associated with the universal properties of $F(A)$ and $F^{ab}(A)$, respectively, and let $i: A \to G$ be the inclusion map. Then $\varphi$ is the unique homomorphism such that $\varphi \circ j = i$. There is also a unique homomorphism $\psi: F^{ab}(A) \to G$ such that $\psi \circ k = i$, and a unique homomorphism $\chi: F(A) \to F^{ab}(A)$ such that $\chi \circ j = k$. This all follows from universal properties.

To verify the authors statement, we must show that $\varphi$ and $\psi$ have the same image. It follows from the above that $i = \psi \circ \chi \circ j$, and hence that $\psi \circ \chi = \varphi$ (by the uniqueness property of $\varphi$). Thus $$\varphi(F(A)) \subset \psi(F^{ab}(A)).$$

But how do I show the reverse inclusion? I suppose showing that $\chi$ is surjective would do the trick, but I have not been able to.

Answer 1

Here is a proof that $\chi$ is surjective that only requires the universal properties of the free and free abelian groups, and the notion of quotient of an abelian group.

Following the given notation: $(F(A),j)$ is a free group on $A$, with $j\colon A\to F(A)$ the structure function. $(F^{\rm ab}(A),k)$ is a free abelian group on $A$, with $k\colon A\to F^{\rm ab}(A)$ the structure function. The map $k\colon A\to F^{\rm ab}(A)$ induces the morphism $\chi\colon F(A)\to F^{\rm ab}(A)$, which is the unique morphism such that $\chi\circ j = k$.

Now consider the abelian group $N = F^{\rm ab}(A)/\chi(F(A))$; we can do the quotient because (i) $\chi$ is a homomorphism, so $\chi(F(A))$ is a subgroup of $F^{\rm ab}(A)$; and (ii) $F^{\rm ab}(A)$ is abelian, so all subgroups are normal. Moreover, because the quotient of an abelian group is abelian, $N$ is abelian.

Let $\pi\colon F^{\rm ab}(A) \to N$ be the canonical morphism onto the quotient, and let $\zeta\colon F^{\rm ab}(A)\to N$ be the map sending everything to the identity. I claim that $\pi\circ k = \zeta\circ k$ as functions from $A$ to $N$. Indeed, we have that for all $a\in A$, $$k(a) = \chi\circ j(a) = \chi(j(a))\subseteq \chi(j(A))\subseteq \chi(F(A)),$$ so $\pi(k(a)) = k(a)\chi(F(A)) = e\chi(F(A))$, the identity element of $N$. On the other hand, $\zeta\circ k(a) = e\chi(F(A))$ because everything in $F^{\rm ab}(A)$ is mapped to the identity under $\zeta$. Thus, $\pi\circ k$ and $\zeta\circ k$ have the same domain (namely $A$), same codomain, and same value at each element of $A$, so they are equal.

The universal property of $F^{\rm ab}(A)$ states that given any function $g\colon A\to N$, there exists a unique morphism $\eta\colon F^{\rm ab}(A)\to N$ such that $g = \eta\circ k$. Taking $g=\zeta\circ k$, we see that $\eta=\zeta$ is such a mophism; but we just saw that $\zeta\circ k = \pi\circ k$, so $\pi$ also works. (Alternatively, the same argument using $\pi\circ k$ shows $\pi$ works, and the fact that $\zeta\circ k = \pi\circ k$ shows they both work for the same function.)

By the uniqueness clause of the universal property, we conclude that $\pi=\zeta$. But that means that the canonical map onto the quotient $F^{\rm ab}(A) \to F^{\rm ab}(A)/\chi(F(A))$ is the same as the zero map. That means that $\chi(F(A))=F^{\rm ab}(A)$, proving that $\chi$ is surjective, as desired.

Answer 2

The universal property of Abelianisation tells you immediately $\varphi$ would factor through $(F(A))^{ab}$; it is an easy theorem (left adjoints compose, really) that this is isomorphic with $F^{ab}(A)$.

Let $\chi:H\to H^{ab}$ be the Abelianisation homomorphism; you care for the case $H=F(A)$, and you want $\chi$ to be surjective because this would entail the image of $\varphi$ is the image of $F^{ab}(A)\to G$, so the replacement is genuine.

But $\chi$ is an epimorphism in the category of groups! Indeed, the uniqueness part of the universal property tells you immediately that if $f\chi=g\chi=:h$, then $f,g$ are two factorisations of $h$ through $\chi$ and must, thus, be equal.

So the only bit of concrete group theory we need, rather than category theory, is the fact that every epimorphism of groups is set-theoretically surjective. This is answered very nicely here.

Answer 3

For whatever it's worth, here follows an argument for the surjectivity of $\chi$ based on the "concrete" constructions of $F(A)$ and $F_{ab}(A)$ given in Aluffi's book.

In the text $F(A)$ is realized as the set of reduced words on $A$, with multiplication defined as juxtaposition followed by reduction, while $F_{ab}(A)$ is realized as $\mathbb Z^{\oplus A}$. (See the book for details, such as how $j$ and $k$ are defined). It is straightforward to design a surjective homomorphism $\chi: F(A) \to F_{ab}(A)$ such that $\chi \circ j = k$. Essentially, let $\chi$ count the number of occurences of each symbol in a word, with "inverse" symbols counted negative.

It remains only to show that this result generalizes to arbitrary realizations of the free groups. Hence let $(j', F'(A))$ and $(k', F'_{ab}(A))$ be some alternate realizations, with induced homomorphism $\chi': F'(A) \to F'_{ab}(A)$ such that $\chi' \circ j' = k'$. We will show that $\chi'$ is surjective too.

Now, $(j, F(A))$ and $(j', F'(A))$ are both initial objects in the category used to define free groups on $A$ (see Aluffi's text for details). Hence they are isomorphic as objects (Proposition 5.4 in the book). Carefully parsing what this means in the present category, we find that the induced homomorphism $\sigma: F'(A) \to F(A)$ is an isomorphism. Similarly, the induced homomorphism $\tau: F'_{ab}(A) \to F_{ab}(A)$ is an isomorphism.

Note that $$k = \tau \circ k' = \tau \circ \chi' \circ j' = \tau \circ \chi' \circ \sigma^{-1} \circ j.$$ By the uniqueness property of $\chi$ we have $\tau \circ \chi' \circ \sigma^{-1} = \chi$, or $$\chi ' = \tau^{-1} \circ \chi \circ \sigma.$$ The right-hand side is a composition of surjective maps, thus $\chi'$ is surjective.

Answer 4

Here is a completely different argument. It is based on an alternative characterization of the subgroup of $G$ generated by $A$, namely as the intersection of all subgroups of $G$ containing $A$.

Lemma: Let $F(A)$ be (any realization of) the free group on $A$ and let $j: A \to F(A)$ be the associated map. Moreover, let $i$ be the inclusion $A \hookrightarrow G$ and let $\varphi: F(A) \to G$ be the unique homomorphism such that $\varphi \circ j = i$. Then the image of $\varphi$ is the intersection of all subgroups of $G$ containing $H$.

Proof: Let $I$ be the intersection of all subgroups of $G$ containing $H$. It is clear that $I \subset \operatorname{im} \varphi$, since $\operatorname{im} \varphi$ is a subgroup of $G$ containing $A$. To prove the reverse inclusion, let $H$ be an arbitrary subgroup of $G$ containing $A$. Let $\psi: F(A) \to H$ be the unique homomorphism such that $\psi \circ j = i'$, where $i': A \hookrightarrow H$. Denoting the inclusion $H \hookrightarrow G$ by $i''$, we have $$i = i'' \circ i' = i'' \circ \psi \circ j,$$ and thus $i'' \circ \psi = \varphi$, by the uniqueness property of $\varphi$. It follows that $$\operatorname{im} \varphi = \operatorname{im} \psi \subset H.$$ Since $H$ was arbitrary we have $\operatorname{im} \varphi \subset I$, which completes the proof.

Corollary: Aluffi's definition of $\langle A \rangle$ does not depend on which realization of $F(A)$ is used.

Finally, when $G$ is abelian, we can replace $F(A)$ by $F^{ab}(A)$ in the above lemma; every step of the proof still works. Hence the definitions of $\langle A \rangle$ via $F(A)$, via $F^{ab}(A)$, and as the intersection of all subgroups containing $A$, are all equivalent.