Confusion on the definition of presentation of Groups

abstract-algebrafree-groupsgroup-theory

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?

Best Answer

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 $q(r_i) = 1_{F/N}$.

Related Question