Definition of Presentation of Group in Dummit’s Abstract Algebra

Question

A subset $S$ of elements of a group $G$ with the property that every element of $G$ can be written as a (finite) product of elements of $S$ and their inverses is called a set of generators of $G$. We shall indicate this notationally by writing $G=\langle S\rangle$ and say $G$ is generated by $S$ or $S$ generates $G$.

Any equations in a general group $G$ that the generators satisfy are called relations in $G$. Thus in $D_{2n}$ we have relations: $r^n=1$, $s^2 = 1$ and $rs = sr^{-1}$. Moreover, in $D_{2n}$ these three relations have the additional property that any other relation between elements of the group may be derived from these three (this is not immediately obvious; it follows from the fact that we can determine exactly when two group elements are equal by using only these three relations).

In general, if some group $G$ is generated by a subset $S$ and there is some collection of relations, say $R_1,R_2,…,R_m$ (here each $R_i$ is an equation in the elements from $S\cup \{ 1 \}$) such that any relation among the elements of $S$ can be deduced from these, we shall call these generators and relations a presentation of $G$ and write $$G= \langle S\mid R_1,R_2,…,R_m \rangle$$

I want to write definition of generator of $G$ in notation format. Let $S\subseteq G$. Then $G=\langle S\rangle$ if $\forall g\in G$, $\exists a_1,…,a_n\in S$ such that $n\in \Bbb{N}$ and $g=\prod_{i=1}^nb_i$, where $b_i\in \{a_i,a_i^{-1}\}$. Note we can take $a_i$ more than once. Is my definition correct?

Question: (1) What is equation in definition of relation? In definition it is said that “generator satisfy” but $r^{-1}\notin S=\{r,s\}$. Though $rs=sr^{-1}$ is equivalent to $(rs)^2=1$.

(2) How to rigorously show that any relation between elements of group may be derived from $r^n=1$, $s^2=1$ and $rs=sr^{-1}$? I think we need to define relation more concretely to prove it.

(3) What does author mean by “we can determine exactly when two group elements are equal by using only these three relations”? Two element are equal when they are equal.

(4) $G=\langle S\rangle$ and $\exists R_1,…,R_m$ such that any relation among the elements of $S$ can be deduced from $R_1,…,R_m$. What if $m=\infty$?

Answer 1

The basic idea of a free object with relations is a bit obscured by viewing it specifically as a notion for groups. In fact, the same construction can work for every algebraic structure, and even with an infinite number of generators and relations.

You start with identifying which operations and which laws you have on your algebraic structure. In the case of groups, you have a binary operation $\cdot$, a unary operation $\,^{-1}$ and a nullary operation $e$, together with the laws $(x\cdot y)\cdot z=x\cdot (y\cdot z)$, $x\cdot e=x$, $e\cdot x=x$, $x\cdot x^{-1}=e$ and $x^{-1}\cdot x=e$.

Now, take your set of generators (say $r,s$) and make all the possible finite expressions, using the operations and brackets. You would get expressions such as $(r\cdot s)^{-1}\cdot((r^{-1}\cdot e)\cdot r^{-1})$ etc.

Now make an algebraic structure of the set of all the expressions (call that set $E$). That structure will have the same operations (with the same arity), except that all the operations will be formal. For example, the result of applying $\,^{-1}$ to the expression $r^{-1}\cdot s$ is merely $(r^{-1}\cdot s)^{-1}$, the result of applying $\cdot$ to, say, $r^{-1}$ and $s\cdot r$ is $r^{-1}\cdot(s\cdot r)$, and the result of the operation $e$ is just the expression $e$.

Fine. Now you have an algebraic structure with the same three operations, on the set of all expressions $E$, but what laws does it satisfy? Probably none. For example, $e\cdot x$ is just $e\cdot x$, which is not the same as $x$.

In the next step, we define what we want to be true (and no more than that!). Recall that a binary relation on $E$ is a subset of $E\times E$, and some binary relations are congruences (i.e. equivalence relations that agree with the operations). This amounts, in our particular story, to the following laws being true (for a relation $R\subseteq E\times E$):

• Reflexivity: All $(w, w)\in R$
• Symmetry: If $(v, w)\in R$, then $(w, v)\in R$
• Transitivity: If $(u, v)\in R$ and $(v, w)\in R$, then $(u, w)\in R$.
• Agreement with $\cdot$: If $(v_1, w_1)\in R$ and $(v_2, w_2)\in R$, then $(v_1\cdot v_2, w_1\cdot w_2)\in R$
• Agreement with $\,^{-1}$: If $(v, w)\in R$, then $(v^{-1}, w^{-1})\in R$.

So, create a set of statements you want to be true, let that be your initial relation $R\subset E\times E$. You will put into $R$:

• For each group law and every variable that there is in it, a bunch of pairs from $E\times E$. For example, for the law $e\cdot x=x$ you would put pairs such as $(e\cdot w, w)$ for all $w\in E$. Or, for the associative law, you would put pairs such as $((u\cdot v)\cdot w,u\cdot(v\cdot w))$ for all $u,v,w\in E$.
• Also put all the relations (don't confuse this with the notion of the binary relation/equivalence/congruence) above) that you want. For example, for the dihedral group, your relations are $r^n=e, s^2=e, rs=sr^{-1}$, so you would put the following three additional pairs: $(\underbrace{(\cdots((r\cdot r)\cdot r)\cdots r)}_{n\text{ r's}}, e), (s\cdot s, e), (r\cdot s, s\cdot r^{-1})$

Now, as the last step: notice that:

• An intersection of any set of congruences on $E$ is again a congruence on $E$.

• There is one trivial congruence on $E$: the full relation $E\times E$ (i.e. any pair $(v,w), v,w\in E$ belongs to it).

Thus, we can take all the congruences on $E$ that contain $R$ and make the (set-theoretic) intersection of all of them. What you get is the smallest congruence on $E$ containing $R$. Call that congruence $\overline{R}$.

Finally, create the quotient structure $E/\overline{R}$. That is your group with given relations.

Of course this should now be followed by a massive proof that:

• What we've got is indeed a group.
• It satisfies the relations given, where the role of $r$ is taken by $r/\overline{R}$ - i.e. by the equivalence class of the expression $r$. (And similar with $s$ and $e$.) For example $(s/\overline{R})\cdot(s/\overline{R})=(s\cdot s)/\overline{R}=e/\overline{R}$ The middle equality is valid because $(s\cdot s, e)\in R\subseteq\overline{R}$.
• Any other group generated by two elements $r, s$ which satisfy the given relations is a homomorphic image of $E/\overline{R}$, i.e. $E/\overline{R}$ is in some sense the "biggest" (or "freest") of all the groups with those relations.