Comparing 2 universal properties.

Question

Here is the first universal property and the map required to prove it:

And here is the second universal property:

My questions are:

1- Are they the same or not? I am guessing that both of them are universal properties for the quotient but why $A$ needed to be Abelian in the second and how this is compensated in the first?

2- Is the map used for proving the first can also be used for proving the second?

Could anyone help me in answering those questions please?

Answer 1

The two are similar, in that the second is a special case of the first, with the added information about the fact that the subgroup $[G,G]$ is connected to abelian groups (in some perhaps mysterious way). The map you get in the second instance is indeed the map that you would have in the first example, if $[G,G]$ plays the role of $H$. It just so happens that anytime you have a map whose kernel contains $[G,G]$, the image of the map will be abelian.

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Where these “universal properties” come from... I’m going to borrow the presentation framework from George Bergman’s An Invitation to General Algebra and Universal Constructions, as I’m teaching a course out of it right now.

Fix your group $G$, and a subgroup $H$ (not necessarily normal). Let us consider all the morphism $f\colon G\to K$ to groups that have that property that $f(h)=f(e)$ for all $h\in H$ (that is, $H\subseteq \mathrm{ker}(f)$). There are many reasons you may be interested in such maps.

Note that if we have such a morphism $f$, and $g\colon K\to M$ is any group morphism from $K$ to another group, then we will automatically get a new morphism from $G$, namely $g\circ f\colon G\to M$. This map will also satisfy that $(g\circ f)(h)=(g\circ f)(e)$ for all $h\in H$, since $g$ must send $e_K$ to $e_M$.

So one may wonder if there is a “universal” example of a map $p$ from $G$ to a group $\mathfrak{G}$, with the property that (i) $p(h)=p(e)$ for all $h\in H$; and (ii) any map $f\colon G\to K$ with $f(h)=f(e)$ for all $h\in H$ can be obtained by composing $p$ with a map from $\mathfrak{G}$ to $K$. That is, we would like a pair $(\mathfrak{G},p)$ with the following universal property:

1. $p\colon G\to \mathfrak{G}$ is a morphism;
2. $p(h)=p(e)$ for all $h\in H$; and
3. If $f\colon G\to K$ is any morphism into a group such that $f(h)=f(e)$ for all $h\in H$, then there exists a unique morphism $g\colon \mathfrak{G}\to K$ such that $f = g\circ p$.

The answer is “yes”; such a “universal pair” is given by taking $\mathfrak{G}=G/N$, where $N$ is the normal closure of $H$ in $G$; and $p\colon G\to G/N$. is the canonical projection, $p(x)=xN$. Then the map $g\colon G/N\to K$ is given by $g(xN) = f(x)$.

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Now let us consider a related problem: say we are interested in maps from $G$ to abelian groups. Again, if $f\colon G\to A$ is a map from $G$ to an abelian group, then any morphism $g\colon A\to B$ of abelian groups will give rise to a new morphism from $G$ to an abelian group by composition, $g\circ f\colon G\to B$. Is there a “universal pair” $(A,p)$ such that

1. $A$ is an abelian group and $p\colon G\to A$ is a morphism.
2. If $B$ is any abelian group and $f\colon G\to B$ is a morphism, then there exists a unique morphism $g\colon A\to B$ such that $f=g\circ p$.

Again, the answer is “yes”: the universal pair is the group $G/[G,G]$ and $p\colon G\to G/[G,G]$ is the projection onto the quotient. One may be led to this group by noting that any map $p\colon G\to A$ form $G$ to an abelian group must satisfy $$p(xy)=p(yx)$$ and hence that $e = p(yx)^{-1}p(xy) = p(x^{-1}y^{-1}xy)$. That is, the elements of $[G,G]$ must lie in the kernel of any such map; this already tells us from the first universal property that the maps will factor through $G/[G,G]$. And since this group is abelian, it must be the “universal example” of a map into an abelian group. That is, we can justify the fact that $(G/[G,G],p)$ is a universal pair into abelian groups by using the first universal property of the quotient.

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In fact, the latter can be viewed in the context of categories: you have the collection of all abelian groups sitting inside the collection of all groups. Moreover, any “abelian group” morphism between abelian groups is a morphism of them as groups, but more importantly, any group group morphism between them is also what one might call an “abelian group morphism”. In the language of category theory, the collection of all abelian groups is a “full subcategory” of the category of abelian groups.

But in fact there is more: the existence of a “universal example” of a map from an arbitrary group to an abelian group says that we have a way to go from “group” to “abelian group” with the following property:

If $G$ is a group and $A$ is an abelian group, then there is a natural bijection between the group morphisms from $G$ to $A$, viewing $A$ as a group, and the abelian group morphism from $G/[G,G]$ to $A$. That is, $$\mathbf{Group}(G,i(A)) \cong \mathbf{Ab}(G/[G,G],A).$$ where $i(A)$ means that we view $A$ as a group, “forgetting” that it is in fact an abelian group.

This is an instance of something called an “adjunction”. It is a key concept in category theory, especially as it relates to algebras. We can view the abelianization $G/[G,G]$ as the realization of this adjunction. One can also view the first universal property as an instance of an adjunction, though the category is more complicated and perhaps looks more unnatural, so I won’t go into them (especially if you are not very familiar with Category Theory).