Defining Groups via the Universal Property

Question

The snippets below on group presentation led to the following questions:

1. What is the explanation/intuition of the universal property. Basically what (2) is saying. I understand free groups (though not how to write a good definition), generators, and quotient groups at a high level, but not the formal definition universal property and how you define a group by the universal property. Which is part of the next question.
2. What "every $ϕ : A \to H$ can be [extended to] $ϕ^\ast : G \to H$" means. In (2), the definition of universal property. Basically some help with the definition of universal property. I don't understand what "extended to a unique homomorphism" means. I understand homomorphism though.
3. Where the quotient $G \simeq F(X)/\ker(ψ)$ came from. Specifically, I don't understand how (4) and (5) were derived.
4. At a high level, how to describe a group via generators and relators. Based off the universal property (as described in (3)).

From there I think I could figure out how to generate a group.

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This is my inaccurate attempt at writing a definition for the free group :/

(1) Basically this is my understanding of free groups which leads up to group presentations... Given a group and a subset of the group, the group is called "free" if its elements are entirely composed of combinations of the subset. The subset is called the "generating set". Formally, given a group subset $A \subset G$, it's denoted: $$\langle A\rangle = \{a^{\epsilon_1}_{i_1}, \dotsc , a^{\epsilon_n}_{i_n} \mid a_{i_j} \in A, \epsilon _j \in \{1, −1\}, n \in N\}$$

(2) Then I get a bit hazy here. The universal property for a group and generating set is a set of maps from the generating set to a new subset that can be ~extended to a unique homomorphism~ from the group to that second subset. Formally, every $\phi : A \to H$ can be extended to $\phi^\ast : G \to H$.

Then this doc states:

(3) Defining various free objects (groups, rings, etc.) via their universal properties is a standard way to define universal objects in category theory... The universal property of free groups allows one to describe arbitrary groups in terms of generators and relators.

That's what I'd like to do.

Next, it talks about group presentations (the crux of the question):

(4) Let $G$ be a group with a generating set $A$. By the universal property of free groups there exists a homomorphism $ψ : F(X) \to G$ such that $ψ(a) = a$ for $a \in A$. It follows that $ψ$ is onto [unclear], so by the first isomorphism theorem $G \simeq F(X)/\ker(ψ)$. In this event $ker(ψ)$ is viewed as the set of relators of $G$, and a group word $w \in ker(ψ)$ is called a relator of $G$ in generators $A$.

Wikipedia has a different but related phrasing:

(5) Formally, the group $G$ is said to have the above presentation [$\langle S\mid R\rangle$] if it is isomorphic to the quotient of a free group on $S$ by the normal subgroup generated by the relations $R$.

I understand $F(X)/\ker(ψ)$ is a quotient, and the $\ker$ comes from the first isomorphism theorem, where the kernel of $ψ$ is a normal subgroup of $G$. So the two definitions line up it seems.

Answer 1

What "every $ϕ : A \to H$ can be [extended to] $ϕ^\ast : G \to H$" means. In (2), the definition of universal property. Basically some help with the definition of universal property. I don't understand what "extended to a unique homomorphism" means. I understand homomorphism though.

Well if $G,H$ are groups and $A\subseteq G$ is a subset (not necessarily a subgroup) and $f:A\to H$ is a function then extending $f$ to $G$ simply means finding a function $f':G\to H$ such that $f'(g)=f(g)$ for $g\in A$.

If that extension is a homomorphism then we say that $f$ can be extended to a homomorphism. Note that $f$ need not be a homomorphism. Indeed, if $A$ is not a subgroup then it doesn't even make sense to talk about homomorphisms $A\to H$.

If there is only one such $f'$ extending $f$ then we say that this extension is unique.

Note that not every function can be extended to a homomorphism. Consider $f:\{0\}\to\mathbb{Z}$, $f(0)=1$. This function cannot be extended to a homomorphism $\mathbb{Z}\to\mathbb{Z}$, because every homomorphism maps the neutral element to the neutral element. Similarly $f:\{1, -1\}\to\mathbb{Z}$, $f(1)=1$, $f(-1)=0$. Do you know why?

But some can. For example if $f:\mathbb{Z}\to G$ is a group homomorphism then for $n\in\mathbb{N}$

$$f(n)=f(1+\cdots +1)=f(1)\cdots f(1)=f(1)^n$$

Analogously $f(-n)=f(1)^{-n}$ and so $f$ only depends on the value $f(1)$. We can look at this property from the other side: every function $h:\{1\}\to G$ can be uniquely extended to a homomorphism $h':\mathbb{Z}\to G$ by the formula:

$$h'(n)=h(1)^n$$

In particular $\mathbb{Z}$ is the free group over an alphabet composed of only one element, i.e. $\mathbb{Z}\simeq F(\{1\})$.

Where the quotient $G \simeq F(X)/\ker(ψ)$ came from.

Let $G$ be a group. Consider a set of generators $X\subseteq G$. Now take the function $f:X\to G$, $f(x)=x$. By the universal property it can be extended to a unique homomorphism $\psi:F(X)\to G$. Note that we treat $X$ as a subset of $F(X)$ since obviously elements of $X$ represent words of length $1$.

By the universal property of free groups there exists a homomorphism $\psi : F(X) \to G$ such that $\psi(a) = a$ for $a \in A$. It follows that $\psi$ is onto [unclear]

I assume it's a typo and it should be $F(A)$. Anyway the easiest way to see that $\psi$ is onto (i.e. surjective) is simply if you take $A=G$.

For a proper set of generators $A\subseteq G$ you can go through the definition of $F(A)$ and see that that every possible combination of generators can be realized in $F(A)$ and thus in the image of $\psi$. So as long as $A\subseteq G$ is a set of generators then $\psi$ is onto.

Formally, the group $G$ is said to have the above presentation [$\langle S\mid R\rangle$] if it is isomorphic to the quotient of a free group on $S$ by the normal subgroup generated by the relations $R$.

The notion $\langle S\ |\ R\rangle$ simply means

$$\langle S\ |\ R\rangle := F(S)/N(R)$$

where $N(R)$ is the smallest normal subgroup of $F(S)$ containing $R$. In particular it is assumed that $R\subseteq F(S)$. For example

$$\langle x, y\ |\ x^2, y^2, xyx^{-1}y^{-1}\rangle\simeq\mathbb{Z}_2\oplus\mathbb{Z}_2$$ $$\langle x, y\ |\ x^3, y^2, xyx^{-2}y\rangle\simeq S_3$$

What is the explanation/intuition of the universal property.

Pretty much the same as in linear algebra: every homomorphism $H:F(X)\to G$ is uniquely determined by values on $X$ and any values are valid. So $X$ is a basis for $F(X)$.

Answer 2

Definition (1) isn't good. Here are some better alternatives:

(1') Fix a set S. The free group on $S$ is the group $X$ consisting of sequences $w = (s_1^{\epsilon_1}, \dots, s_n^{\epsilon_n})$ for $s_i\in S$ and $\epsilon_i = \pm 1$, modulo the relations \begin{align*} (\dots, s_n, \sigma^{+1}, \sigma^{-1}, s_{n+1}, \dots) &= (\dots, s_n, s_{n+1}, \dots); \\ (\dots, s_n, \sigma^{-1}, \sigma^{+1}, s_{n+1}, \dots) &= (\dots, s_n, s_{n+1}, \dots). \end{align*} The group multiplication is given by $$(s_1, \dots, s_n)(s'_1, \dots, s'_m) = (s_1, \dots, s_n)(s'_1, \dots, s'_m).$$

(1'') Let $X$ be a group, and let $S\subset X$ be a subset. Then $X$ is free on $S$ if for every function of sets $f:S \to G$ with $G$ a group, there exists a unique group homomorphism $\tilde f:X \to G$ with $\tilde f\vert S = f$.

(1''') The forgetful functor $Gp \to Set$ admits a left adjoint $\cal{F}$. The free group on $S$ is ${\cal F}(S)$.

(I would warn you that (1') is a bad definition; (1'') and (1''') are much better. The only real point of (1') is that it proves that free groups exist, but there are better ways of doing that.)

The point is that we can now define groups in terms of presentations. Assume that the free groups $\cal{F}(S)$ on an arbitrary (nonempty) set $S$ exists. Let $X$ be a group, and let $A\subset X$ be a generating set. (We could take $A = X$, but in practice we want $A$ to be finite and as small as possible.) Applying the universality property in (1'') to the inclusion $A \to X$ shows that there exists a map of groups $f:{\cal F}(A) \to X$ whose image contains $A$ and is thus surjective. Hence $X = {\cal F}(A)/(\ker f)$.