Free Products and Amalgamation

Given: The objects and hypotheses in the statement.

For groups as in def-group, a syllable is a tagged pair $(i,g)$ with $i\in I$ and $g\in G_i\setminus\{e_i\}$. A reduced syllable word is a finite list of syllables, indexed by a natural length as in def-natural-numbers, in which adjacent tags differ. The empty list is allowed. At a concatenation seam, adjacent syllables from the same factor are multiplied and an identity result is deleted; this elementary reduction is repeated until the seam is reduced. (Reduced syllable words in a family of groups).

For every set $X$, the triple $(\operatorname{Sym}(X), \circ, \mathrm{id}_X)$ of def-symmetric-group is a group (def-group); the inverse of a permutation $f$ is its inverse function $f^{-1}$. If $X$ contains three distinct elements $a$, $b$, $c$, then $\operatorname{Sym}(X)$ is not abelian: the transpositions $\tau = (a\,b)$ and $\rho = (b\,c)$ satisfy $\tau \circ \rho \ne \rho \circ \tau$. ($\operatorname{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

Let $(M,\cdot,e)$ and $(M',\cdot',e')$ be monoids (def-semigroup-and-monoid). A monoid homomorphism from $M$ to $M'$ is a function $f : M \to M'$ such that - (H1) $f(x \cdot y) = f(x) \cdot' f(y)$ for all $x, y \in M$; - (H2) $f(e) = e'$. Let $G$ and $G'$ be groups (def-group). A group homomorphism from $G$ to $G'$ is a function $f : G \to G'$ satisfying (H1) alone: $f(xy) \;=\; f(x)\, f(y) \qquad \text{for all } x, y \in G .$ Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f(e) = e'$ and $f(x^{-1}) = f(x)^{-1}$ (lem-group-homomorphism-basic-properties). For monoids it does not follow and must be assumed, which is why the two definitions differ. A homomorphism from a structure to itself is an endomorphism. The identity map of $M$ is a monoid homomorphism, and a composite of monoid homomorphisms is one, since $(g \circ f)(xy) = g(f(x)f(y)) = g(f(x))\,g(f(y))$ and $(g \circ f)(e) = g(e') = e''$; the same computation, without the second clause, shows a composite of group homomorphisms is a group homomorphism. (Monoid homomorphism and group homomorphism).

Given: The objects and hypotheses in the statement.

For a family $(G_i)_{i\in I}$, a free product is a group $F$ with homomorphisms $\iota_i:G_i\to F$ in the sense of def-group-homomorphism, such that for every group $H$ and every family of homomorphisms $f_i:G_i\to H$, there is a unique homomorphism $f:F\to H$ satisfying $f\circ\iota_i=f_i$ for all $i$. It is denoted $\ast_{i\in I}G_i$. Injectivity of the maps $\iota_i$ is not part of this definition. (The free product of an arbitrary family of groups).

For groups as in def-group, a syllable is a tagged pair $(i,g)$ with $i\in I$ and $g\in G_i\setminus\{e_i\}$. A reduced syllable word is a finite list of syllables, indexed by a natural length as in def-natural-numbers, in which adjacent tags differ. The empty list is allowed. At a concatenation seam, adjacent syllables from the same factor are multiplied and an identity result is deleted; this elementary reduction is repeated until the seam is reduced. (Reduced syllable words in a family of groups).

For each $i\in I$ and $g\in G_i$, left multiplication at the first syllable defines a permutation $P_{i,g}$ of the set of reduced words. One has $P_{i,g}^{-1}=P_{i,g^{-1}}$ and $P_{i,gh}=P_{i,g}\circ P_{i,h}$, so $g\mapsto P_{i,g}$ is a group homomorphism. (Factor elements act by mutually inverse permutations on reduced syllable words).

Let $(M,\cdot,e)$ and $(M',\cdot',e')$ be monoids (def-semigroup-and-monoid). A monoid homomorphism from $M$ to $M'$ is a function $f : M \to M'$ such that - (H1) $f(x \cdot y) = f(x) \cdot' f(y)$ for all $x, y \in M$; - (H2) $f(e) = e'$. Let $G$ and $G'$ be groups (def-group). A group homomorphism from $G$ to $G'$ is a function $f : G \to G'$ satisfying (H1) alone: $f(xy) \;=\; f(x)\, f(y) \qquad \text{for all } x, y \in G .$ Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f(e) = e'$ and $f(x^{-1}) = f(x)^{-1}$ (lem-group-homomorphism-basic-properties). For monoids it does not follow and must be assumed, which is why the two definitions differ. A homomorphism from a structure to itself is an endomorphism. The identity map of $M$ is a monoid homomorphism, and a composite of monoid homomorphisms is one, since $(g \circ f)(xy) = g(f(x)f(y)) = g(f(x))\,g(f(y))$ and $(g \circ f)(e) = g(e') = e''$; the same computation, without the second clause, shows a composite of group homomorphisms is a group homomorphism. (Monoid homomorphism and group homomorphism).

Given: The objects and hypotheses in the statement.

The reduced syllable words in $(G_i)_{i\in I}$ form a group under concatenation followed by seam reduction. The one-syllable maps $G_i\to W$ make this group a free product of the family. (Reduced syllable words form the free product of a family of groups).

For groups as in def-group, a syllable is a tagged pair $(i,g)$ with $i\in I$ and $g\in G_i\setminus\{e_i\}$. A reduced syllable word is a finite list of syllables, indexed by a natural length as in def-natural-numbers, in which adjacent tags differ. The empty list is allowed. At a concatenation seam, adjacent syllables from the same factor are multiplied and an identity result is deleted; this elementary reduction is repeated until the seam is reduced. (Reduced syllable words in a family of groups).

For a family $(G_i)_{i\in I}$, a free product is a group $F$ with homomorphisms $\iota_i:G_i\to F$ in the sense of def-group-homomorphism, such that for every group $H$ and every family of homomorphisms $f_i:G_i\to H$, there is a unique homomorphism $f:F\to H$ satisfying $f\circ\iota_i=f_i$ for all $i$. It is denoted $\ast_{i\in I}G_i$. Injectivity of the maps $\iota_i$ is not part of this definition. (The free product of an arbitrary family of groups).

Given: The objects and hypotheses in the statement.

For groups as in def-group, a syllable is a tagged pair $(i,g)$ with $i\in I$ and $g\in G_i\setminus\{e_i\}$. A reduced syllable word is a finite list of syllables, indexed by a natural length as in def-natural-numbers, in which adjacent tags differ. The empty list is allowed. At a concatenation seam, adjacent syllables from the same factor are multiplied and an identity result is deleted; this elementary reduction is repeated until the seam is reduced. (Reduced syllable words in a family of groups).

Every element of $\ast_{i\in I}G_i$ has a unique reduced syllable expression. The identity is represented by the empty word, and no nonempty reduced word represents the identity. (Normal form theorem for free products).

The order of a finite group. Let $G$ be a group (def-group) whose underlying set is finite (def-countable), so that $G \approx n$ for some $n \in \mathbb{N}$ (def-equinumerous). That natural number is unique: if $G \approx n$ and $G \approx n'$ then $n \approx n'$, since $\approx$ is symmetric and transitive, and then $n = n'$ by claim 3 of lem-pigeonhole. The order of $G$ is that unique natural number, written $|G|$. A group is infinite when its underlying set is not finite, and $|G|$ is then not defined. The order of an element. Let $G$ be any group and $g \in G$, with natural powers as in def-group-power. Put $S_g \;:=\; \{\, k \in \mathbb{N} \;:\; k \ge 1 \text{ and } g^{k} = e \,\} \;\subseteq\; \mathbb{N}.$ - If $S_g \ne \varnothing$, the order of $g$ is its least element, $\operatorname{ord}(g) \;:=\; \min S_g \;\in\; \mathbb{N},$ which exists by the well-ordering principle (thm-well-ordering-principle): every nonempty subset of $\mathbb{N}$ has a least element, and that element is unique, being $\le$ every element of $S_g$ and a member of it. We then say $g$ has finite order. - If $S_g = \varnothing$ we say $g$ has infinite order and write $\operatorname{ord}(g) = \infty$, where $\infty$ is a symbol reserved for this case and is not a natural number. No arithmetic is performed with it here. By construction $\operatorname{ord}(g) \ge 1$ whenever it is finite, and $\operatorname{ord}(g) = 1$ exactly when $g = e$, since $g^{1} = g$. Every element of a finite group has finite order. If $G$ is finite then $S_g \ne \varnothing$ for every $g \in G$, by lem-order-of-element-exists, so $\operatorname{ord}(g)$ is a natural number. (The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity).

Let $G$ be a group (def-group) with identity $e$, let $g, h \in G$, and let powers be as in def-group-power. For all $m, n \in \mathbb{Z}$: 1. $g^{m+n} = g^{m} g^{n}$; 2. $g^{-m} = (g^{m})^{-1}$; 3. $(g^{m})^{n} = g^{mn}$; 4. $g^{m} g^{n} = g^{n} g^{m}$: any two powers of one element commute; 5. if $gh = hg$ then $(gh)^{n} = g^{n} h^{n}$. Claim 5 is false in general without its hypothesis: in a group in which $g$ and $h$ do not commute the equation can fail already at $n = 2$, and a witness is recorded on the companion page. Claims 1 and 3 hold in any monoid (def-semigroup-and-monoid) for exponents in $\mathbb{N}$, and so does claim 5 for exponents in $\mathbb{N}$ under the same commuting hypothesis; only the extension to negative exponents needs inverses. (Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$, and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

Given: The objects and hypotheses in the statement.

Let $G$ be a group (def-group). The center of $G$ is $Z(G):=\{z\in G:zg=gz\text{ for every }g\in G\}.$ Thus $Z(G)$ consists of the elements that commute with every element of $G$. Its subgroup and normality properties are proved in lem-center-is-normal. (The center $Z(G)$ of a group).

For groups as in def-group, a syllable is a tagged pair $(i,g)$ with $i\in I$ and $g\in G_i\setminus\{e_i\}$. A reduced syllable word is a finite list of syllables, indexed by a natural length as in def-natural-numbers, in which adjacent tags differ. The empty list is allowed. At a concatenation seam, adjacent syllables from the same factor are multiplied and an identity result is deleted; this elementary reduction is repeated until the seam is reduced. (Reduced syllable words in a family of groups).

Every element of $\ast_{i\in I}G_i$ has a unique reduced syllable expression. The identity is represented by the empty word, and no nonempty reduced word represents the identity. (Normal form theorem for free products).

Given: The objects and hypotheses in the statement.

Every element of $\ast_{i\in I}G_i$ has a unique reduced syllable expression. The identity is represented by the empty word, and no nonempty reduced word represents the identity. (Normal form theorem for free products).

A group homomorphism is injective if and only if its kernel is trivial. For a group homomorphism $f:G\to H$, $f$ is injective exactly when $\ker f=\{e_G\}$. (A group homomorphism is injective if and only if its kernel is trivial).

Given: The objects and hypotheses in the statement.

For a family $(G_i)_{i\in I}$, a free product is a group $F$ with homomorphisms $\iota_i:G_i\to F$ in the sense of def-group-homomorphism, such that for every group $H$ and every family of homomorphisms $f_i:G_i\to H$, there is a unique homomorphism $f:F\to H$ satisfying $f\circ\iota_i=f_i$ for all $i$. It is denoted $\ast_{i\in I}G_i$. Injectivity of the maps $\iota_i$ is not part of this definition. (The free product of an arbitrary family of groups).

Group isomorphisms, automorphisms and the set $\operatorname{Aut}(G)$. An isomorphism $f:G\to H$ is a bijective group homomorphism (def-group-homomorphism, def-injection-surjection-bijection). When $G=H$, it is an automorphism of $G$. Write $\operatorname{Aut}(G):=\{f:G\to G:f\text{ is an automorphism}\}.$ (Group isomorphisms, automorphisms and the set $\operatorname{Aut}(G)$).

Given: The objects and hypotheses in the statement.

For a family $(G_i)_{i\in I}$, a free product is a group $F$ with homomorphisms $\iota_i:G_i\to F$ in the sense of def-group-homomorphism, such that for every group $H$ and every family of homomorphisms $f_i:G_i\to H$, there is a unique homomorphism $f:F\to H$ satisfying $f\circ\iota_i=f_i$ for all $i$. It is denoted $\ast_{i\in I}G_i$. Injectivity of the maps $\iota_i$ is not part of this definition. (The free product of an arbitrary family of groups).

Every canonical factor homomorphism $\iota_i:G_i\to\ast_jG_j$ is injective. (Every canonical factor map into a free product is injective).

Let $(M,\cdot,e)$ and $(M',\cdot',e')$ be monoids (def-semigroup-and-monoid). A monoid homomorphism from $M$ to $M'$ is a function $f : M \to M'$ such that - (H1) $f(x \cdot y) = f(x) \cdot' f(y)$ for all $x, y \in M$; - (H2) $f(e) = e'$. Let $G$ and $G'$ be groups (def-group). A group homomorphism from $G$ to $G'$ is a function $f : G \to G'$ satisfying (H1) alone: $f(xy) \;=\; f(x)\, f(y) \qquad \text{for all } x, y \in G .$ Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f(e) = e'$ and $f(x^{-1}) = f(x)^{-1}$ (lem-group-homomorphism-basic-properties). For monoids it does not follow and must be assumed, which is why the two definitions differ. A homomorphism from a structure to itself is an endomorphism. The identity map of $M$ is a monoid homomorphism, and a composite of monoid homomorphisms is one, since $(g \circ f)(xy) = g(f(x)f(y)) = g(f(x))\,g(f(y))$ and $(g \circ f)(e) = g(e') = e''$; the same computation, without the second clause, shows a composite of group homomorphisms is a group homomorphism. (Monoid homomorphism and group homomorphism).

Given: The objects and hypotheses in the statement.

Let $F(X)$ be a free group and let $R\subseteq F(X)$ be a set of words, called relations. The group with presentation $\langle X\mid R\rangle:=F(X)/\langle\!\langle R\rangle\!\rangle_{F(X)}$ is the quotient by the normal closure of $R$. The members of $X$ are its generators. In this quotient, every relation in $R$ becomes the identity, as do all consequences forced by normality. (Group presentation by generators and relations).

Let $\langle X\mid R\rangle$ be a presentation, let $H$ be a group, and let $u:X\to H$ be a function. If the evaluation of every $r\in R$ under $u$ is $e_H$, then there is a unique homomorphism $\overline u:\langle X\mid R\rangle\longrightarrow H$ with $\overline u([x])=u(x)$ for every $x\in X$. Moreover, $\overline u$ is surjective if and only if $u(X)$ generates $H$. (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group).

For a family $(G_i)_{i\in I}$, a free product is a group $F$ with homomorphisms $\iota_i:G_i\to F$ in the sense of def-group-homomorphism, such that for every group $H$ and every family of homomorphisms $f_i:G_i\to H$, there is a unique homomorphism $f:F\to H$ satisfying $f\circ\iota_i=f_i$ for all $i$. It is denoted $\ast_{i\in I}G_i$. Injectivity of the maps $\iota_i$ is not part of this definition. (The free product of an arbitrary family of groups).

Any two free products of the same family are connected by a unique isomorphism commuting with every canonical factor map. (Free products are unique up to a unique factor-compatible isomorphism).

In a presentation $\langle X\mid R\rangle$ as in def-group-presentation, an element $r\in R\subseteq F(X)$ is called a defining relator. The equation $r=1$ that it imposes in the quotient is a defining relation. More generally, an equation $u=v$ may be recorded by the relator $u^{-1}v$. The published definition uses the common looser convention of calling the members of $R$ relations; both conventions define the same quotient group. A presentation is finitely generated when $X$ is finite, finitely related when $R$ is finite, and finite when both $X$ and $R$ are finite. A group is called finitely generated, finitely related, or finitely presented when it admits a presentation with the corresponding property. For finitely generated groups this agrees with generation by a finite subset in the sense of def-generated-subgroup. (Relators and relations; finitely generated, finitely related, and finite presentations).

Given: The objects and hypotheses in the statement.

A free group on a set $X$ is a group $F(X)$ together with a map $i:X\to F(X)$ such that, for every group $G$ and every function $u:X\to G$, there is a unique group homomorphism $\widehat u:F(X)\to G$ satisfying $\widehat u\circ i=u.$ The reduced-word construction supplies such a group; the construction and its universal property are established in thm-reduced-words-form-the-free-group. When no ambiguity arises, $x\in X$ is identified with its image $i(x)$. (Free group on a set of generators).

If $(F,i)$ and $(F',i')$ are free groups on the same set $X$, then there is a unique group isomorphism $\phi:F\to F'$ such that $\phi\circ i=i'.$ (Free groups on the same set are uniquely isomorphic compatibly with their generators).

For a family $(G_i)_{i\in I}$, a free product is a group $F$ with homomorphisms $\iota_i:G_i\to F$ in the sense of def-group-homomorphism, such that for every group $H$ and every family of homomorphisms $f_i:G_i\to H$, there is a unique homomorphism $f:F\to H$ satisfying $f\circ\iota_i=f_i$ for all $i$. It is denoted $\ast_{i\in I}G_i$. Injectivity of the maps $\iota_i$ is not part of this definition. (The free product of an arbitrary family of groups).

Any two free products of the same family are connected by a unique isomorphism commuting with every canonical factor map. (Free products are unique up to a unique factor-compatible isomorphism).

Given: The objects and hypotheses in the statement.

For pairwise disjoint sets $X_i$, the free product of the free groups $F(X_i)$ is a free group on $\bigsqcup_iX_i$. (Free groups on disjoint bases freely multiply to the free group on their union).

A free group on a set $X$ is a group $F(X)$ together with a map $i:X\to F(X)$ such that, for every group $G$ and every function $u:X\to G$, there is a unique group homomorphism $\widehat u:F(X)\to G$ satisfying $\widehat u\circ i=u.$ The reduced-word construction supplies such a group; the construction and its universal property are established in thm-reduced-words-form-the-free-group. When no ambiguity arises, $x\in X$ is identified with its image $i(x)$. (Free group on a set of generators).

Let $G$ be a group and $g \in G$, with integer powers as in def-group-power. Then $\langle g \rangle \;=\; \{\, g^{n} \;:\; n \in \mathbb{Z} \,\} ,$ the cyclic subgroup generated by $g$ (def-generated-subgroup) being exactly the set of integer powers of $g$. Consequently every cyclic group is abelian, and so is every cyclic subgroup of any group. ($\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, and every cyclic group is abelian).

Let $G$ be a group, $g \in G$, and let orders be as in def-order-in-a-group. Throughout, a natural number written where an integer is expected means its image under the embedding $\iota : \mathbb{N} \to \mathbb{Z}$ of lem-nat-embeds-int. Finite order. Suppose $\operatorname{ord}(g) = n$ with $n \in \mathbb{N}$, $n \ge 1$. Then: 1. for every $k \in \mathbb{Z}$, $g^{k} = e$ if and only if $k = qn$ for some $q \in \mathbb{Z}$, that is, if and only if $n \mid k$ (thm-division-algorithm-in-z); 2. the powers $g^{0}, g^{1}, \dots, g^{n-1}$ are pairwise distinct: if $i, j \in \mathbb{N}$ with $i < n$, $j < n$ and $g^{i} = g^{j}$, then $i = j$; 3. $\langle g \rangle = \{\, g^{s} : s \in \mathbb{N},\ s < n \,\}$ and $\langle g \rangle \approx n$; so $\langle g \rangle$ is finite with $|\langle g \rangle| = n = \operatorname{ord}(g)$. Infinite order. If $\operatorname{ord}(g) = \infty$ then for $j, k \in \mathbb{Z}$, $g^{j} = g^{k}$ implies $j = k$; so the integer powers of $g$ are pairwise distinct and $\langle g \rangle$ is not finite. (If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$).

Let $G$ be a group (def-group) with identity $e$, let $g, h \in G$, and let powers be as in def-group-power. For all $m, n \in \mathbb{Z}$: 1. $g^{m+n} = g^{m} g^{n}$; 2. $g^{-m} = (g^{m})^{-1}$; 3. $(g^{m})^{n} = g^{mn}$; 4. $g^{m} g^{n} = g^{n} g^{m}$: any two powers of one element commute; 5. if $gh = hg$ then $(gh)^{n} = g^{n} h^{n}$. Claim 5 is false in general without its hypothesis: in a group in which $g$ and $h$ do not commute the equation can fail already at $n = 2$, and a witness is recorded on the companion page. Claims 1 and 3 hold in any monoid (def-semigroup-and-monoid) for exponents in $\mathbb{N}$, and so does claim 5 for exponents in $\mathbb{N}$ under the same commuting hypothesis; only the extension to negative exponents needs inverses. (Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$, and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

Given: The objects and hypotheses in the statement.

A free product of a family of infinite cyclic groups is a free group on one chosen generator from each factor. The empty family gives the free group on the empty set. (A free product of copies of the infinite cyclic group is a free group).

A free group $F$ has finite rank if it admits a finite free basis. In that case its rank is $\operatorname{rank}(F):=|B|,$ where $B$ is any finite free basis of $F$. This is well-defined by thm-finite-free-bases-have-the-same-cardinality. This definition is deliberately restricted to free groups that admit a finite free basis. It neither defines rank for a free group whose bases are infinite nor asserts that arbitrary infinite free bases have the same cardinality. (The rank of a free group admitting a finite basis).

Given: The objects and hypotheses in the statement.

Given homomorphisms $f:K\to G$ and $h:K\to H$ as in def-group-homomorphism, a pushout is a group $P$ with homomorphisms $i_G:G\to P$ and $i_H:H\to P$ such that $i_G\circ f=i_H\circ h$, and such that every compatible pair $u:G\to Q$, $v:H\to Q$ factors through a unique $w:P\to Q$ with $w\circ i_G=u$ and $w\circ i_H=v$. The maps $f,h$ need not be injective. (Pushouts of group homomorphisms).

For a family $(G_i)_{i\in I}$, a free product is a group $F$ with homomorphisms $\iota_i:G_i\to F$ in the sense of def-group-homomorphism, such that for every group $H$ and every family of homomorphisms $f_i:G_i\to H$, there is a unique homomorphism $f:F\to H$ satisfying $f\circ\iota_i=f_i$ for all $i$. It is denoted $\ast_{i\in I}G_i$. Injectivity of the maps $\iota_i$ is not part of this definition. (The free product of an arbitrary family of groups).