Confusion on the definition of presentation of Groups

Question

We define a group presentation to be an ordered pair denoted by $\langle S | R \rangle$ where $S$ is an arbitrary set and $R$ is an arbitrary subset of the free group $F(S)$. We call the elements of $S$ the generators and the elements of $R$ the relators.

The group presentation defines a group (also denoted by $\langle S | R \rangle$) as the following quotient group $$\langle S | R \rangle = F(S) / \overline{R}$$ where $\overline{R}$ is the normal closure of $R$ in $F(S)$. Now in the book I'm reading Introduction to Topological Manifolds by John Lee the following is stated

"Each of the relators $r \in R$ represents a particular product of powers of the generators that is equal to $1$ in the quotient"

Now if I understand correctly $1$ in the quotient is just simply $$1_{F(S)/\bar{R}} = 1_F(S) \cdot \overline{R}$$ which equals $\overline{R}$ set-theoretically. And $r$ representing a particular product of powers of the generators means $$r = s_1^{n_1} \cdot \dots s_k^{n_k}$$

I don't see how $r$ can equal $1$ in the quotient (which is $1_F(S) \cdot \overline{R}$) when it isn't even a coset to begin with. What exactly does the author mean when he said the above?

Answer 1

Almost all textbooks give terrible definitions of group given by generators and relations, and "clarifications" of what is meant are usually even worse than their absense.

Here we have free group $F$ and normal group $N$ — whichever way you define it, it's normal closure of $\{r_i\}$. So there's canonical homomorphism $q: F \to F/N$. Quoted statement is just a terrible way to say that $q(r_i) = 1_{F/N}$.

Answer 2

He says that $$\overline{r}=\overline{s}_1^{n_1}\cdot\dots\cdot \overline{s}_k^{n_k}=\overline{1},$$ where a bar means an element of the quotient group (so $\overline{1}=1_{F(S)/\overline{R}}$ in your notation above).

Answer 3

The key word is "represents".

I much prefer thinking about quotient sets in general as being sets equipped with a different notion of equality. (Formally, this is known as a setoid.) Given a normal subgroup $N\triangleleft G$, we can define an equivalence relation on (the carrier set of) $G$ as $g\sim h \iff gh^{-1}\in N$. (You should prove that this is indeed a congruence, i.e. it's an equivalence relation [reflexive, symmetric, transitive], and it respects the operations of a group.) We can recover $N$ from this congruence via $N = \{g\in G\mid g\sim 1\}$. In fact, given any congruence on $G$, the equivalence class of $1$ is a normal subgroup of $G$. (Prove this.) Thus normal subgroups of $G$ are in one-to-one correspondence with congruences on $G$. This is why normal subgroups are important.

Now we can view $G/N$ as being $G$ except that we use $\sim$ for equality. That is, we can imagine that when we say $g=h$ as elements of $G/N$, we mean $g\sim h$.¹ With this perspective, every element of $N$ is "equal" to $1$ in $G/N$.

In a more traditional perspective, $G/N$ is the set of equivalence classes of the congruence mentioned before. These are the cosets. A representative of an equivalence class is simply an element of that equivalence class. Given such a representative, $h$, we can immediately recover the equivalence class via $\{g\in G\mid g\sim h\}$. Any element of an equivalence class represents that equivalence class. If we write $[h]$ for $\{g\in G\mid g\sim h\}$, then $[g]=[h]\iff g\sim h$. In group theory, $[g]$ is typically written $gN$, i.e. as a coset. So what the quote is saying from this perspective is that $[r] = [s_1^{n_1}s_2^{n_2}\dots s_k^{n_k}] = [s_1^{n_1}][s_1^{n_2}]\dots[s_k^{n_k}] = [1]$ for $r\in\overline R$.

It is pretty common to fluidly move between working in terms of representatives and their equivalence classes. Working in terms of representatives requires you to ensure that everything you do doesn't depend on the specific representative of the equivalence class you've chosen but is usually clearer (both conceptually and notationally).

¹ In a usual set-theoretic context, this can only be an intuition though you could simply choose to talk about $\sim$. In other foundations, this perspective can be made literally true.