A group presentation turns symbols and equations into a group. This chapter constructs that process before applying it to braids. Fix an integer \(n\geq3\). Here \(A_n\) means the abstract group defined by the Artin presentation; its identification with moving points is the theorem proved in Chapter 5. After the equivalences are established, \(B_n\) is the common name for the braid group in any of its three models.
Prerequisites: groups, homomorphisms, and subgroups
A group is a set with an associative multiplication, an identity \(1\), and an inverse for each element. A homomorphism \(\phi:G\to H\) satisfies \(\phi(gh)=\phi(g)\phi(h)\). A subgroup is a subset that contains \(1\) and is closed under products and inverses. The subgroup generated by a subset consists of all finite products of its elements and their inverses, including the empty product \(1\).
The kernel \(\ker\phi=\{g:\phi(g)=1\}\) is a subgroup and satisfies \(g(\ker\phi)g^{-1}=\ker\phi\): apply \(\phi\) to \(gkg^{-1}\). A subgroup with this conjugation property is called normal. See Alphabeta Math's groups and subgroups and normal subgroups and quotients.
2.1. Words and reduction
Free groups retain exactly the cancellations forced by inverses. Additional equations will be imposed only after this construction.
Definition 2.1. Words and elementary reduction
Let \(S\) be a set of symbols. Make a disjoint set of formal inverse symbols \(S^{-1}\), with \((s^{-1})^{-1}=s\). A word is a finite string of letters in \(S\sqcup S^{-1}\). The empty word is denoted \(\varepsilon\); concatenation puts one word after another. If \(w=\ell_1\cdots\ell_k\), its inverse word is \(w^{-1}=\ell_k^{-1}\cdots\ell_1^{-1}\).
An elementary cancellation deletes an adjacent pair \(\ell\ell^{-1}\). A word is reduced if no such pair occurs. Define \(u\sim v\) when finitely many cancellations or reverse insertions transform \(u\) into \(v\). These moves may occur inside any surrounding words, so \(u\sim v\) implies \(aub\sim avb\).
Theorem 2.2. Reduced normal form
Every word is equivalent to a unique reduced word. Any sequence of elementary cancellations terminates at that word.
Proof.
Scan a word from left to right, keeping a stack of letters. When the next letter is inverse to the stack's last letter, remove that last letter; otherwise append the new letter. Write \(R(w)\) for the final stack. It is reduced, and each scan step is either concatenation or one allowed cancellation, so \(R(w)\sim w\).
For a letter \(\ell\), let \(T_\ell\) be the operation of processing \(\ell\) on a reduced stack. Its inverse is \(T_{\ell^{-1}}\). If \(\ell\) was appended, the next operation removes it. If \(\ell\) removed a terminal \(\ell^{-1}\), the next operation restores that letter: the remaining stack cannot end in \(\ell\), because the original stack was reduced. Thus processing an inserted pair \(\ell\ell^{-1}\) has no effect, even after a nonempty prefix. Processing the remaining suffix then gives the same result. Hence \(R\) is unchanged by every allowed move.
Every reduced word \(r\) satisfies \(R(r)=r\). Consequently, if two reduced words are equivalent, both equal the same \(R(w)\). Repeated cancellation terminates because length decreases, and its endpoint must be this unique reduced word. Compare the reduced-word theorem in Alphabeta Math.
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JavaScript enables the cancellation demonstration.
2.2. Free groups
Definition 2.3. The free group on a set
Let \(S\) be a set. The free group \(F(S)\) is the set of word classes under elementary insertion and cancellation, with multiplication \([u][v]=[uv]\), identity \([\varepsilon]\), and inverse \([w]^{-1}=[w^{-1}]\).
Proposition 2.4. Group structure
The operations in Definition 2.3 are well defined and make \(F(S)\) a group. The map \(S\to F(S)\), \(s\mapsto[s]\), is injective.
Proof.
Compatibility with surrounding words makes this well-defined. Concatenation is associative, so this multiplication is associative. The identity is \([\varepsilon]\). Successively cancelling from the middle of \(ww^{-1}\) and \(w^{-1}w\) shows that \([w^{-1}]\) is the inverse of \([w]\). Thus \(F(S)\) is a group, and its elements may be represented uniquely by reduced words.
The symbols \(s\in S\) give distinct elements \([s]\), since distinct one-letter words are reduced. We identify each symbol with its class when no ambiguity results. This group is free on \(S\) in the following precise sense:
□Theorem 2.5. Universal property of the free group
For every group \(H\) and every function \(f:S\to H\), there is exactly one homomorphism \(\widehat f:F(S)\to H\) extending \(f\).
Proof.
\[ \widehat f([s_1^{\epsilon_1}\cdots s_k^{\epsilon_k}]) = f(s_1)^{\epsilon_1}\cdots f(s_k)^{\epsilon_k}, \qquad \epsilon_j\in\{1,-1\}. \]A cancellation or insertion evaluates to multiplying by \(f(s)f(s)^{-1}=1\), so the formula is well-defined. Concatenation evaluates to multiplication, which proves the homomorphism property. Every word is a product of generator classes and their inverses, so any extending homomorphism must have this formula. This also proves uniqueness. “Free” therefore means that assigning the generators requires no compatibility conditions beyond those already built into a group.
□2.3. Normal closures and quotients
Suppose we want words \(u\) and \(v\) to become equal. In a group this means making \(uv^{-1}=1\). Such a word is called a relator. If a relator becomes the identity, its inverse and every conjugate \(wrw^{-1}\) must also become the identity, as must products of those elements. Killing just a subgroup generated by relators would not generally account for conjugation.
Definition 2.6. Normal closure
For \(\mathcal R\subseteq F(S)\), its normal closure \(\langle\!\langle\mathcal R\rangle\!\rangle\) is the smallest normal subgroup of \(F(S)\) containing \(\mathcal R\).
Worked example for Definition 2.6
Counterexample: A subgroup generated by relators need not be normal
Lemma 2.7. Products of conjugates
The normal closure of \(\mathcal R\) consists exactly of the empty product and the products
\[ (w_1r_1^{\epsilon_1}w_1^{-1})\cdots (w_kr_k^{\epsilon_k}w_k^{-1}), \qquad r_j\in\mathcal R,\quad w_j\in F(S),\quad \epsilon_j\in\{1,-1\}. \]Counterexample: A subgroup generated by relators need not be normal
Proof.
Include the empty product. Products stay in this set, inverses reverse the factors and change their signs, and conjugating a factor by \(g\) replaces \(w_j\) by \(gw_j\). The set is therefore a normal subgroup containing \(\mathcal R\). Every normal subgroup containing \(\mathcal R\) contains all these factors and products, proving minimality. See also normal closure as products of conjugates.
□Definition 2.8. Cosets and the quotient set
If \(N\trianglelefteq G\), define \(gN=\{gn:n\in N\}\) and \(G/N=\{gN:g\in G\}\). Two representatives define the same coset precisely when their quotient lies in \(N\).
Alphabeta Math: cosets · quotients
Worked example for Definition 2.8
Counterexample: Coset multiplication fails without normality
Proposition 2.9. Quotient group
For \(N\trianglelefteq G\), the rule below makes \(G/N\) a group. The projection \(q(g)=gN\) is a surjective homomorphism with kernel \(N\).
\[ (gN)(hN)=ghN. \]Worked example for Proposition 2.9
Counterexample: Coset multiplication fails without normality
Proof.
Cosets satisfy \(gN=hN\) if and only if \(h=gn\) for some \(n\in N\). If the cosets are equal, \(h\in gN\); conversely, \(h=gn\) gives \(hN=gnN=gN\). Thus the coset equality criterion used in the definition follows from the subgroup laws.
If \(g'=gn\) and \(h'=hm\), then \(g'h'=gh(h^{-1}nh)m\). Normality puts \(h^{-1}nh\) in \(N\), so the product does not depend on the representatives. Associativity, identity \(N\), and inverse \(g^{-1}N\) now follow from the corresponding laws in \(G\). The projection \(q:G\to G/N\), \(q(g)=gN\), is a surjective homomorphism with kernel \(N\).
□Proposition 2.10. Universal property of the quotient
If \(\phi:G\to H\) is a homomorphism and \(N\subseteq\ker\phi\), there is a unique homomorphism \(\bar\phi:G/N\to H\) satisfying \(\bar\phi\circ q=\phi\).
Proof.
Moreover, if \(\phi:G\to H\) kills \(N\), there is exactly one homomorphism \(\bar\phi:G/N\to H\) with \(\bar\phi(gN)=\phi(g)\). This is well-defined because \(\phi(gn)=\phi(g)\); the formula also proves uniqueness and multiplicativity. This is the quotient's universal property.
□2.4. Group presentations
Imposing a relator also imposes its conjugates and products. The normal closure records exactly these consequences.
Definition 2.11. Group presentation
A presentation defines the group
\[ \langle S\mid\mathcal R\rangle :=F(S)/\langle\!\langle\mathcal R\rangle\!\rangle. \]Writing an equation \(u=v\) in the relation list means including the relator \(uv^{-1}\). Two words represent the same element precisely when \(uv^{-1}\in\langle\!\langle\mathcal R\rangle\!\rangle\): indeed \(uN=vN\) is equivalent to \(v^{-1}u\in N\), and normality makes this equivalent to \(uv^{-1}\in N\). See Alphabeta Math's definition of a group presentation.
Proposition: equality as a finite sequence of relation moves
Proposition 2.12. Relation moves
Two words define the same element of a presented group if and only if free insertions, cancellations, and replacements by defining equations connect them.
Proof.
Besides free insertion and cancellation, allow replacing either side of a defining equation by the other inside any word. Every move preserves its quotient element. Conversely, if a word is a product of conjugates of relators and their inverses, choose word representatives for its factors. The original word and this expanded product have the same reduced word, so free insertions and cancellations connect them. A relator \(uv^{-1}\) becomes \(vv^{-1}\) by replacing \(u\) with \(v\), then cancels; its inverse \(vu^{-1}\) cancels in the same way. Each conjugate therefore becomes the empty word. The normal-closure criterion proves that every word equal to the identity can be transformed to the empty word by finitely many such moves. Applying this to \(uv^{-1}\), and appending \(v\), proves the corresponding statement for equality of any two words. Insertions and reverse relation moves are essential; a delete-only procedure need not succeed.
□Theorem 2.13. Homomorphisms from a presentation
An assignment \(S\to H\) extends uniquely to a homomorphism \(\langle S\mid\mathcal R\rangle\to H\) if and only if every relator evaluates to the identity.
Proof.
To define a homomorphism from \(\langle S\mid\mathcal R\rangle\) to \(H\), assign each generator an element of \(H\) and check that every relator evaluates to \(1\). The free-group property gives a homomorphism on \(F(S)\). Its kernel is normal and contains \(\mathcal R\), hence contains the normal closure; the quotient property then gives the desired homomorphism. Conversely, any homomorphism out of the presented group must respect those equations.
□Example 2.14. A cyclic presentation
For \(m\geq1\), \(\langle a\mid a^m=1\rangle\cong\mathbb Z/m\mathbb Z\).
Proof.
A reduced word on one generator is \(a^k\), and multiplication adds exponents. The normal closure of \(a^m\) is the subgroup of powers with exponent divisible by \(m\), so quotienting identifies exponents modulo \(m\). This illustrates why a presentation asserts both the listed equations and exactly their group-theoretic consequences.
□2.5. The Artin braid group
Definition 2.15. Artin braid group
Take \(S=\{s_1,\ldots,s_{n-1}\}\). The abstract Artin group \(A_n\) is
\[ A_n=\left\langle s_1,\ldots,s_{n-1}\ \middle|\, \begin{array}{ll} s_is_j=s_js_i & (1\leq i,j\leq n-1,\ |i-j|\geq2),\\[3pt] s_is_{i+1}s_i =s_{i+1}s_is_{i+1} & (1\leq i\leq n-2) \end{array}\right\rangle. \]These are the distant commutation and adjacent braid relations. This is a definition at this stage, matching the Artin presentation entry. For a standard research survey of the algebraic and geometric definitions, see Birman and Brendle, Braids: A Survey, §1.2, equation (2).
Remark 2.16. Relators and notation
For complete precision, the relator set contains
\[ s_is_js_i^{-1}s_j^{-1} \quad (|i-j|\geq2), \]and
\[ s_is_{i+1}s_i s_{i+1}^{-1}s_i^{-1}s_{i+1}^{-1} \quad (1\leq i\leq n-2). \]The symbols \(s_i\) first denote free generators and then their images in the quotient; we reserve \(\sigma_i\) for the configuration loops in chapter 3. For \(n=3\), there are two generators and just the adjacent relation \(s_1s_2s_1=s_2s_1s_2\). For \(n=4\), the new distant relation is \(s_1s_3=s_3s_1\).
Both sides of an adjacent braid relation are distinct reduced free words. The quotient makes them equal. Thus free reduction alone does not give a unique representative for a braid. A presentation provides a criterion for equality through normal closure, but it does not by itself supply an efficient equality algorithm.
Worked example for Remark 2.16
Counterexample: Neither free reduction nor the two simple invariants decide braid equality
2.6. Permutations and exponent sum
Proposition 2.17. Two quotient homomorphisms
There are surjective homomorphisms \(p:A_n\to S_n\) and \(e:A_n\to\mathbb Z\) determined by \(p(s_i)=(i\ i+1)\) and \(e(s_i)=1\). Each \(s_i\) has infinite order.
Alphabeta Math: permutation map · exponent sum
Worked example for Proposition 2.17
Counterexample: Neither free reduction nor the two simple invariants decide braid equality
Proof.
Permutation map. Let \(S_n\) be the group of permutations of \(\{1,\ldots,n\}\), with ordinary function composition. Set \(t_i=(i\ i+1)\). Disjoint transpositions commute. On the three affected entries, both \(t_it_{i+1}t_i\) and \(t_{i+1}t_it_{i+1}\) exchange \(i\) and \(i+2\) and fix \(i+1\); they fix every other entry. The presentation property therefore gives a homomorphism
\[ p:A_n\longrightarrow S_n,\qquad p(s_i)=t_i. \]It is surjective: repeatedly swapping adjacent out-of-order entries sorts any list of distinct numbers, so adjacent transpositions generate every permutation. The equations \(t_i^2=1\) in the target are extra equations, not part of the defining braid relations. This algebraic map uses usual composition; a chronological record of labelled trajectories requires an explicit convention, as discussed in Chapter 3.
Exponent sum. Assign every \(s_i\) the integer \(1\) in the additive group \(\mathbb Z\). Both sides of each relation have the same sum, so there is a homomorphism
\[ e:A_n\longrightarrow\mathbb Z,\qquad e(s_{i_1}^{\epsilon_1}\cdots s_{i_k}^{\epsilon_k}) =\sum_{j=1}^{k}\epsilon_j. \]It follows that every \(s_i\) has infinite order: if \(s_i^m=1\), applying \(e\) gives \(m=0\). In particular \(s_i^2\neq1\), although \(p(s_i^2)=1\). Exponent sum is only a partial test: \(s_1s_2^{-1}\) has sum zero but is nontrivial, because its image \(t_1t_2\) is nonidentity.
Surjectivity of \(e\) follows from \(e(s_1)=1\).
□2.7. Comparison with geometric braids
Eventually the formal generator \(s_i\) will correspond to the configuration-loop class \([\sigma_i]\), a counterclockwise exchange of the \(i\)-th and \((i+1)\)-st points in a real-axis base configuration: the left point travels below and the right point above. That interpretation is a candidate correspondence, not a consequence of the symbols alone. Showing that geometric exchanges satisfy the relations constructs a homomorphism; showing that there are no additional geometric relations proves that it is an isomorphism. Chapter 5 addresses both tasks after Chapter 3 defines configuration-space braids. Chapter 4 introduces the third model, and Chapter 6 compares it with moving points.
Sources and further reading
- Joan S. Birman and Tara E. Brendle, Braids: A Survey, §§1.1–1.3: the classical definitions; §1.2, equation (2): the Artin presentation.
- Alphabeta Math: reduced words form the free group, normal closure as products of conjugates, and group presentations. The proofs needed here are given above.
- Alphabeta Math: the braid group by Artin presentation.
Worked examples
These calculations and proofs use this series’ chronological loop products, inverse-endpoint mapping classes, and counterclockwise positive point exchanges. Related statements link to their examples.
Worked example 1. Reduction, inversion, and the free-group operations
Applies to: Definition 2.1, Theorem 2.2, Definition 2.3, Proposition 2.4.
Proof and calculation.
On \(S=\{a,b\}\), take \(w=ab b^{-1}a^{-1}b\). Cancelling \(bb^{-1}\), then \(aa^{-1}\), gives the reduced word \(b\). Initially the only cancellation is \(bb^{-1}\); afterwards the only cancellation is \(aa^{-1}\), so every cancellation sequence gives this result in this example. Its inverse word is \(b^{-1}ab b^{-1}a^{-1}\), reducing to \(b^{-1}\), as required by reversal and inversion. Concatenating the reduced words gives \(bb^{-1}\sim\varepsilon\). The words \(a,b\) are reduced and distinct, so their classes are distinct by the normal-form proof above; generator injection is visible here. Associativity before taking classes is literal concatenation of finite strings, and insertion or deletion inside a surrounding string gives the same equivalence class, checking the quotient operations on this calculation.
Worked example 2. An explicit map out of a free group
Applies to: Theorem 2.5.
Proof and calculation.
Choose \(f(a)=(12),f(b)=(23)\) in \(S_3\), with ordinary right-factor-first composition. Evaluate each word by multiplying these assigned permutations and their inverses. Inserting \(\ell\ell^{-1}\) inserts a permutation times its inverse, so evaluation is constant on word classes. Concatenating words multiplies the evaluations, proving it is a homomorphism. For example \(ab\mapsto(123)\) and \(aba^{-1}\mapsto(13)\). Any extension must have these evaluations because a word is a product of its letters; this proves uniqueness directly for this assignment.
Worked example 3. A normal closure, its cosets, and a quotient factorization
Applies to: Definition 2.6, Lemma 2.7, Definition 2.8, Proposition 2.9, Proposition 2.10.
Proof and calculation.
Let \(G=F(\{a,b\})\), \(r=aba^{-1}b^{-1}\), and \(N=\langle\!\langle r\rangle\!\rangle\). The element \(br^{-1}b^{-1}\) lies in \(N\): normality requires each conjugate of the inverse relator. Products of such conjugates remain in \(N\), including the empty product 1. In \(G/N\), the relation \(abN=baN\) lets every word be rearranged to \(a^m b^nN\), including negative exponents by multiplying the commutation equation by inverses.
Define \(\phi:G\to\mathbb Z^2\) by counting total \(a\)- and \(b\)-exponents. Cancellation changes neither count, and concatenation adds counts. The relator, its conjugates, inverses, and their products all have count \((0,0)\); hence \(N\subseteq\ker\phi\). Consequently \(\bar\phi(gN)=\phi(g)\) is well defined: if \(gN=hN\), then \(h^{-1}g\in N\), so their counts agree. It is a homomorphism, with inverse \((m,n)\mapsto a^m b^nN\); rearrangement proves both composites are identities. Thus these are distinct cosets indexed by \(\mathbb Z^2\), and \((a^m b^nN)(a^p b^qN)=a^{m+p}b^{n+q}N\). Any factorization of \(\phi\) must send \(gN\) to \(\phi(g)\), proving its uniqueness and checking the quotient universal property.
Worked example 4. A relation move with a certificate in the normal closure
Applies to: Definition 2.11, Proposition 2.12, Theorem 2.13.
Proof and calculation.
In \(\langle a,b\mid aba=bab\rangle\), put \(r=aba\,b^{-1}a^{-1}b^{-1}\). The word \(w=aba\,b^{-1}a^{-1}b^{-1}\) becomes empty by replacing its initial \(aba\) with \(bab\) and then cancelling \(bab b^{-1}a^{-1}b^{-1}\). A contextual version \(bwb^{-1}\) is a conjugate of \(r\), so also lies in its normal closure; the same substitution inside the surrounding \(b, b^{-1}\) followed by cancellations proves it trivial. This exhibits both descriptions of a presentation relation.
The assignment \(a\mapsto(12),b\mapsto(23)\) respects \(aba=bab\), since both sides are \((13)\). Word evaluation therefore kills \(r\), and kills any product of its conjugates. It descends to this presentation exactly as the coset formula above, uniquely because \(a,b\) generate. An assignment to an abelian group may instead send \(a\) and \(b\) to the same chosen element: the adjacent relation then evaluates to the same cube. These checks establish the required relator condition, rather than assuming every assignment descends.
Worked example 5. The actual Artin relation ranges and two quotient maps
Applies to: Definition 2.15, Remark 2.16, Proposition 2.17.
Proof and calculation.
For \(n=3\), there are generators \(a=s_1,b=s_2\) and the sole defining equation \(aba=bab\). There is no distant relation: their index difference is 1. For \(n=4\), \(s_1s_3=s_3s_1\) becomes available because the difference is 2. The corresponding adjacent relator is \(aba b^{-1}a^{-1}b^{-1}\), with inverses in the reversed order prescribed by \((bab)^{-1}\).
Sending \(s_i\) to \((i\ i+1)\) satisfies distant relations since disjoint transpositions commute. On \(\{i,i+1,i+2\}\), both adjacent triple products exchange the first and third symbols and fix the middle, so the adjacent relation holds too. Adjacent swaps generate every permutation: move the required first entry to position 1 by successive swaps, then repeat on the remaining positions. This proves surjectivity of \(p\). The exponent-count map is well defined because the relations preserve their sums, is onto because \(e(s_1)=1\), and proves \(s_i\) has infinite order. For \(a b^{-1}\in A_3\), these give \(e=0\) but \(p=(12)(23)=(123)\ne1\), showing a zero exponent sum need not give the identity.
Counterexamples and distinctions
Each example refutes the stated stronger claim while retaining the hypotheses explicitly checked in its proof.
Counterexample 1. A subgroup generated by relators need not be normal
Related statements: Definition 2.6, Lemma 2.7.
Claim refuted. Ordinary subgroup generation can replace normal closure in a presentation.
Proof.
In \(F(a,b)\), the subgroup \(\langle a\rangle\) consists exactly of powers \(a^m\). The reduced word \(bab^{-1}\) has letters \(b\) at its ends and cannot be any such power, by reduced normal form. Thus this subgroup is not normal, although it contains the proposed relator \(a\). Setting \(a=1\) must also make \(bab^{-1}=1\), so a quotient presentation has to kill more than ordinary powers of \(a\).
Counterexample 2. Coset multiplication fails without normality
Related statements: Definition 2.8, Proposition 2.9.
Claim refuted. The rule (gH)(kH)=gkH works for an arbitrary subgroup H.
Proof.
In \(S_3\), let \(H=\{1,(12)\}\). Both 1 and \((12)\) represent the coset \(H\). Multiplying representatives on the right by \((23)\) would yield either \((23)H\) or \((12)(23)H\). These cosets differ: equality would require \((23)^{-1}(12)(23)=(13)\in H\), which is false. Thus the proposed operation depends on the chosen representative. This proves the need for normality with a concrete failure of well-definedness.
Counterexample 3. Neither free reduction nor the two simple invariants decide braid equality
Related statements: Remark 2.16, Proposition 2.17.
Claim refuted. Distinct freely reduced words give different braids, or permutation and exponent sum together determine a braid.
Proof.
The reduced words \(aba\) and \(bab\) are distinct in \(F(a,b)\) but equal in \(A_3\) by its defining relation, refuting the first claim. For the second, use matrices
\[A=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad B=\begin{pmatrix}1&0\\-1&1\end{pmatrix},\qquad ABA=BAB=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.\]Their determinants are 1, so the assignment \(a\mapsto A,b\mapsto B\) defines a homomorphism from \(A_3\): evaluation kills the relator as explicitly computed. Now \(w=a^2b^2a^{-2}b^{-2}\) has exponent sum zero, and permutation identity since each squared generator maps to 1. Yet
\[A^2B^2=\begin{pmatrix}-3&2\\-2&1\end{pmatrix},\quad A^2B^2A^{-2}=\begin{pmatrix}-3&8\\-2&5\end{pmatrix},\quad A^2B^2A^{-2}B^{-2}=\begin{pmatrix}13&8\\8&5\end{pmatrix}\ne I.\]Thus \(w\ne1\). The matrix map need not be injective to prove this nonidentity; a homomorphism always sends the identity to \(I\).
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