Chapter 10

Hard

Free Groups & Presentations

00 · Symbol Glossary

$S$A generating set (just a set, not yet a group)

An arbitrary set of symbols — think of them as letters of an alphabet — that will be used to build a group from nothing but formal combinations of themselves.

$F(S)$Free group on S

The group of all reduced words in $S$ and formal inverses of $S$ , under concatenation. The "most unconstrained" possible group containing $S$ — no relation holds among its elements except what the group axioms force.

$s^{-1}$Formal inverse symbol

For each $s\in S$ , a new formal symbol $s^{-1}$ , distinct from $s$ itself, destined to become $s$ 's actual group inverse once $F(S)$ is built.

$\langle S \mid R\rangle$Group presentation

Read "the group generated by $S$ subject to relations $R$ ." Shorthand for $F(S)/N$ , where $N$ is the smallest normal subgroup containing every word in $R$ .

$w=1$A relation

A single equation, written as a relator word $w$ set equal to the identity. " $aba^{-1}b^{-1}=1$ " is the relation forcing $a$ and $b$ to commute, since it rearranges to $ab=ba$ .

01 · Building a Group From Nothing but Symbols

Every group so far has been built from already-existing mathematical objects: numbers, permutations, matrices. This chapter runs the process in reverse: start with a bare set of symbols and manufacture the least constrained group that could possibly contain them.

Construction — The Free Group F(S)

Given a set $S$ , form a new set of symbols $S^{-1} = \{s^{-1} : s\in S\}$ (one formal inverse per generator). A word is a finite string of symbols from $S\cup S^{-1}$ . A word is reduced if it contains no adjacent pair $s\,s^{-1}$ or $s^{-1}s$ (which would obviously "cancel" once these symbols become genuine group inverses). The free group $F(S)$ is the set of all reduced words (including the empty word, playing the role of $e$ ), with operation: concatenate two words, then repeatedly cancel any adjacent inverse pairs until the result is reduced again.

Well-definedness is a genuine theorem, taken as given here

Checking that this operation is associative — that reducing "eagerly" during concatenation never depends on which adjacent pair you cancel first — is a real theorem (classically proved via van der Waerden's trick: represent each generator as a permutation of the set of all reduced words, and verify associativity there instead, where it's automatic). We take this construction as given and spend this chapter's effort on what makes it useful.

Example — Reducing a word in $F(\{a,b\})$

The word $ab a^{-1} a b^{-1} b$ reduces: the middle $a^{-1}a$ cancels, giving $ab\,b^{-1}b$ ; then $b^{-1}b$ cancels, giving $ab$ . The fully reduced form is $ab$ — a genuinely different (shorter) element of $F(\{a,b\})$ .

Common mistake — Free groups are almost never abelian

In $F(\{a,b\})$ , is $ab=ba$ ? As reduced words, $ab$ and $ba$ are different strings of symbols, with no cancellation available to relate them — they are genuinely different elements. Unless $S$ has at most one element, $F(S)$ is always nonabelian: nothing in the construction forces any two distinct generators to commute, and the "least constrained" philosophy of free groups means nothing is ever assumed beyond the bare group axioms.

02 · The Universal Property

The reduced-word construction is a means to an end. What actually characterizes $F(S)$ — and what makes it useful without ever thinking about cancellation again — is a single mapping property.

Theorem 10.1 — Universal Property of Free Groups

Let $\iota: S\to F(S)$ send each generator to the corresponding length-1 word. For any group $G$ and any function $f: S\to G$ (assigning an arbitrary target element to each generator, with no constraints), there exists a unique homomorphism $\varphi: F(S)\to G$ with $\varphi(\iota(s)) = f(s)$ for every $s\in S$ .

Proof

Existence. For a reduced word $w = s_1^{\epsilon_1}s_2^{\epsilon_2}\cdots s_k^{\epsilon_k}$ (each $s_i\in S$ , each $\epsilon_i=\pm1$ ), define:

\varphi(w) := f(s_1)^{\epsilon_1}f(s_2)^{\epsilon_2}\cdots f(s_k)^{\epsilon_k} \in G

(with $\varphi(\text{empty word}) = e_G$ ). This is a legitimate function of $w$ , since $w$ is already given in reduced form. To see $\varphi$ is a homomorphism, first extend the same formula to unreduced concatenations of words — it clearly satisfies $\varphi(uv)=\varphi(u)\varphi(v)$ for the naive (unreduced) concatenation, by simply combining the two symbol lists. Cancellation of an adjacent pair $s\,s^{-1}$ (or $s^{-1}s$ ) inside a word changes nothing about $\varphi$ 's value, since $f(s)f(s)^{-1}=e_G$ (respectively $f(s)^{-1}f(s)=e_G$ ) contributes nothing to the product. So $\varphi$ takes the same value on a word before and after reduction, and $\varphi(w_1\ast w_2) = \varphi(w_1)\varphi(w_2)$ holds for $F(S)$ 's actual operation $\ast$ (concatenate, then reduce) as well.

Uniqueness. Suppose $\psi: F(S)\to G$ is any homomorphism with $\psi(\iota(s))=f(s)$ for all $s\in S$ . For a reduced word $w=s_1^{\epsilon_1}\cdots s_k^{\epsilon_k}$ , viewed as the product $\iota(s_1)^{\epsilon_1}\cdots\iota(s_k)^{\epsilon_k}$ inside $F(S)$ , the homomorphism property (repeatedly applied) together with Theorem 6.1(b) (homomorphisms preserve inverses, handling the $\epsilon_i=-1$ cases) forces:

\psi(w) = \psi(\iota(s_1))^{\epsilon_1}\cdots\psi(\iota(s_k))^{\epsilon_k} = f(s_1)^{\epsilon_1}\cdots f(s_k)^{\epsilon_k} = \varphi(w)

So $\psi=\varphi$ on every element of $F(S)$ : the extension is forced, hence unique.

Why 'free' — no relation is ever assumed

The universal property says exactly this: to define a homomorphism out of $F(S)$ , you may send the generators absolutely anywhere in the target group, with zero constraints, and the rest of the homomorphism is then completely determined. No other group has this much freedom — any group with even one nontrivial relation among its generators would restrict which target assignments are allowed. This is the precise sense in which $F(S)$ is the "least constrained" group containing $S$ .

03 · Presentations

Most groups are not free — they satisfy extra relations ( $r^n=e$ in $\mathbb{Z}_n$ , $srs=r^{-1}$ in $D_n$ ). A presentation builds exactly such a group by starting from $F(S)$ and forcing chosen words to become the identity.

Definition — Presentation

Given a set $S$ and a set of words $R\subseteq F(S)$ (called relators), let $N\trianglelefteq F(S)$ be the smallest normal subgroup of $F(S)$ containing every element of $R$ (called the normal closure of $R$ — the subgroup generated by all conjugates $gwg^{-1}$ for $w\in R$ , $g\in F(S)$ ; a normal subgroup, by construction, since it's closed under conjugation by every element of $F(S)$ ). The presented group is:

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

Example — Recovering Z_n as a presentation

$\langle a \mid a^n\rangle$ : here $S=\{a\}$ , $R=\{a^n\}$ . Since $F(\{a\})$ is already abelian (a single generator has nothing to fail to commute with), every subgroup is normal (Chapter 05), so $N=\langle a^n\rangle$ (the ordinary cyclic subgroup generated by $a^n$ inside $F(\{a\})\cong\mathbb{Z}$ ). The quotient $F(\{a\})/N \cong \mathbb{Z}/n\mathbb{Z} = \mathbb{Z}_n$ — a presentation recovering exactly the group from Chapter 01, built entirely from the single relation " $a$ has order (dividing) $n$ ."

Example — A presentation for $D_n$

$\langle r,s \mid r^n,\, s^2,\, (sr)^2\rangle$ . The relator $(sr)^2 = srsr$ set equal to the identity rearranges (multiply both sides on the right by $r^{-1}s^{-1}$ , using $s^{-1}=s$ since $s^2=e$ ) to exactly $srs=r^{-1}$ — the relation identified geometrically back in Chapter 02, Section 05.

Example — Forcing commutativity with a relator

$\langle a,b \mid aba^{-1}b^{-1}\rangle$ : the single relator $aba^{-1}b^{-1}$ set to the identity rearranges to $ab=ba$ — forcing exactly one relation, that $a$ and $b$ commute, and nothing more. This presented group turns out to be $\mathbb{Z}\times\mathbb{Z}$ (Chapter 07): free on two commuting generators, with no bound on either generator's order.

04 · The Mapping Property of Presentations

Theorem 10.2 — Von Dyck's Theorem

Suppose $G$ is generated by elements $\{g_s\}_{s\in S}$ (one chosen target per generator), and every relator $w\in R$ becomes the identity of $G$ when each $s$ is replaced by $g_s$ throughout $w$ . Then there is a surjective homomorphism $\bar\varphi: \langle S\mid R\rangle \to G$ sending $sN \mapsto g_s$ .

Proof

By Theorem 10.1 (universal property), the function $f: S\to G$ , $f(s)=g_s$ , extends to a unique homomorphism $\varphi: F(S)\to G$ . By hypothesis, every relator $w\in R$ satisfies $\varphi(w)=e_G$ (substituting $g_s$ for each $s$ is exactly what $\varphi$ computes), so $R\subseteq\ker\varphi$ . By Theorem 6.3, $\ker\varphi\trianglelefteq F(S)$ , and since $N$ is the smallest normal subgroup containing $R$ , $N\subseteq\ker\varphi$ .

Define $\bar\varphi: F(S)/N \to G$ by $\bar\varphi(wN) = \varphi(w)$ . Well-defined: if $w_1N=w_2N$ , then $w_1w_2^{-1}\in N\subseteq\ker\varphi$ , so $\varphi(w_1)=\varphi(w_2)$ (the same style of argument used in Theorem 6.5's First Isomorphism Theorem proof). Homomorphism: inherited directly from $\varphi$ . Surjective: since $\{g_s\}$ generate $G$ , every element of $G$ is a product of the $g_s$ and their inverses, and $\bar\varphi(sN)=\varphi(s)=g_s$ hits each generator, so products of $sN$ 's under $\bar\varphi$ hit every element of $G$ .

Example — D_n really is the group its presentation describes

The genuine rotation $r_0$ and reflection $s_0$ in $D_n$ satisfy $r_0^n=e$ , $s_0^2=e$ , $(s_0r_0)^2=e$ (Chapter 02). By Theorem 10.2, there is a surjection $\bar\varphi: \langle r,s\mid r^n,s^2,(sr)^2\rangle \to D_n$ .

Von Dyck's Theorem only guarantees a surjection, not automatically an isomorphism — a presentation could, in principle, describe a strictly larger group that merely maps onto the target. Showing the surjection above is actually injective requires a separate counting argument: every reduced word in $\langle r,s\mid \ldots\rangle$ can be rewritten, using only the three relations, into the form $r^is^j$ with $0\le i<n$ , $j\in\{0,1\}$ — giving at most $2n$ distinct elements in the presented group. Since $\bar\varphi$ is a surjection onto $D_n$ (which has exactly $2n$ elements), the presented group has at least $2n$ elements too. Combining both bounds forces exactly $2n$ , and a surjection between finite sets of equal size is a bijection: $\bar\varphi$ is an isomorphism, confirming $D_n \cong \langle r,s\mid r^n,s^2,(sr)^2\rangle$ exactly, not merely as a quotient.

Common mistake — Assuming every presentation's surjection is automatically injective

Von Dyck's Theorem by itself never proves a presentation equals a specific target group — only that the target is some quotient of the presented group. Confirming an isomorphism (as in the $D_n$ example) always requires an independent argument bounding the presented group's size from above, matched against the target's known size. Skipping this step is one of the most common errors when working with presentations.

05 · Exercises

EXERCISE 10.1

Concatenate the two words directly, then repeatedly cancel any adjacent symbol-inverse pairs until no more cancellation is possible.

Concatenating $aba^{-1}$ and $ab^{-1}a^{-1}$ gives the string $aba^{-1}ab^{-1}a^{-1}$ . The middle $a^{-1}a$ cancels: $ab\,b^{-1}a^{-1}$ . Then $b\,b^{-1}$ cancels: $aa^{-1}$ . Then $a\,a^{-1}$ cancels: the empty word.

The product is the identity — meaning $ab^{-1}a^{-1}$ is precisely the inverse of $aba^{-1}$ in $F(\{a,b\})$ , consistent with Theorem 2.3's general formula $(xyz)^{-1}=z^{-1}y^{-1}x^{-1}$ applied to $x=a,y=b,z=a^{-1}$ .

In $F(\{a,b\})$ , compute the reduced form of $(aba^{-1})(ab^{-1}a^{-1})$ .

EXERCISE 10.2

Apply the universal property directly: since $F(\{a,b\})$ places no constraints on where generators go, just specify images and read off what $\varphi$ must do to a word.

By Theorem 10.1, define $f(a)=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ , $f(b)=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ (any two elements of $GL_2(\mathbb{R})$ , since the universal property imposes no constraints). This extends uniquely to a homomorphism $\varphi: F(\{a,b\}) \to GL_2(\mathbb{R})$ with $\varphi(a)=f(a)$ , $\varphi(b)=f(b)$ , and, e.g., $\varphi(a^2b^{-1}) = f(a)^2f(b)^{-1}$ , computed by ordinary matrix multiplication.

Using Theorem 10.1, explain how to define a homomorphism from $F(\{a,b\})$ into $GL_2(\mathbb{R})$ , and state what $\varphi(a^2b^{-1})$ must equal.

EXERCISE 10.3

Rearrange the relator $a^2$ set to the identity: what does $a^2=e$ say about the order of $a$ ?

$\langle a \mid a^2\rangle$ : the relation $a^2=e$ means $a$ has order dividing 2. Since $F(\{a\})\cong\mathbb{Z}$ (Chapter 03's infinite cyclic group), quotienting by $\langle a^2\rangle$ gives $\mathbb{Z}/2\mathbb{Z} = \mathbb{Z}_2$ — the same style of computation as the $\mathbb{Z}_n$ example in Section 03, with $n=2$ .

$\langle a \mid a^2\rangle \cong \mathbb{Z}_2$ .

Identify the group $\langle a \mid a^2\rangle$ .

EXERCISE 10.4

Apply Von Dyck's Theorem: does $\mathbb{Z}_4$ have generators satisfying the two given relations?

Try $g=1\in\mathbb{Z}_4$ (additive notation, so " $g^4$ " means $4\cdot1=4\equiv0$ , and " $g^2$ " means $2\cdot1=2$ ). Check: $g^4=0=e_{\mathbb{Z}_4}$ ✓ (satisfies the first relator). But $g^2=2\neq0$ — fails the second relator $a^2=e$ .

Since $\mathbb{Z}_4$ requires an element of order exactly 4 to be generated by a single element (Chapter 03), and any element satisfying both $a^4=e$ and $a^2=e$ has order dividing $\gcd(4,2)=2$ , no generator of $\mathbb{Z}_4$ can satisfy both relations simultaneously. Von Dyck's Theorem requires the relations to hold for the chosen generating elements — here, no valid choice exists, so the theorem does not produce a surjection onto $\mathbb{Z}_4$ for this presentation. (In fact $\langle a\mid a^4,a^2\rangle \cong \mathbb{Z}_2$ , since $a^2=e$ already forces order dividing 2, making $a^4=e$ automatic and redundant.)

Does Von Dyck's Theorem give a surjection from $\langle a \mid a^4, a^2\rangle$ onto $\mathbb{Z}_4$ (using $a\mapsto1$ )? Explain why or why not.

06 · Chapter Summary

Concept	Statement
Free group $F(S)$	Reduced words in $S\cup S^{-1}$ under concatenation-and-cancel
Universal property	Any $f:S\to G$ extends to a unique homomorphism $F(S)\to G$ (Thm 10.1)
Presentation $\langle S\mid R\rangle$	$F(S)/N$ , where $N$ = normal closure of the relators $R$
Von Dyck's Theorem	Generators of $G$ satisfying $R$ give a surjection $\langle S\mid R\rangle \twoheadrightarrow G$ (Thm 10.2)
Presentation vs. isomorphism	Von Dyck gives only a surjection; equality needs an independent size bound
$\mathbb{Z}_n$	$\langle a\mid a^n\rangle$
$D_n$	$\langle r,s\mid r^n,s^2,(sr)^2\rangle$ , confirmed exact via a counting argument
$\mathbb{Z}\times\mathbb{Z}$	$\langle a,b\mid aba^{-1}b^{-1}\rangle$

Next: Chapter 11 — Rings begins the second major arc of the course: structures with two interacting operations, addition and multiplication, generalizing $\mathbb{Z}$ the way groups generalized a single operation.