Free Products and Amalgamation — Examples

Given: The objects and hypotheses in the example.

Suppose each $G_i$ has a presentation $\langle X_i\mid R_i\rangle$, with the alphabets replaced by disjoint copies. Then $\ast_iG_i\cong\left\langle\bigsqcup_iX_i\ \middle|\ \bigcup_iR_i\right\rangle.$ (A free product has the union presentation of presentations of its factors).

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).

For every $n\in\mathbb N$, view $n$ as its canonical nonnegative integer and put $n\mathbb Z:=\{nk:k\in\mathbb Z\}$. Then the left cosets of $n\mathbb Z$ in $(\mathbb Z,+)$ are exactly the congruence classes modulo $n$, and coset addition is the published addition of congruence classes. Thus $(\mathbb Z,+)/n\mathbb Z=(\mathbb Z/n,+)$ as the same group on the same underlying set. This includes $n=0$ and $n=1$. (For every $n\in\mathbb N$, the congruence-class group $(\mathbb Z/n,+)$ is the quotient group $(\mathbb Z,+)/n\mathbb Z$).

Given: The objects and hypotheses in the example.

Suppose each $G_i$ has a presentation $\langle X_i\mid R_i\rangle$, with the alphabets replaced by disjoint copies. Then $\ast_iG_i\cong\left\langle\bigsqcup_iX_i\ \middle|\ \bigcup_iR_i\right\rangle.$ (A free product has the union presentation of presentations of its factors).

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).

For every $n\in\mathbb N$, view $n$ as its canonical nonnegative integer and put $n\mathbb Z:=\{nk:k\in\mathbb Z\}$. Then the left cosets of $n\mathbb Z$ in $(\mathbb Z,+)$ are exactly the congruence classes modulo $n$, and coset addition is the published addition of congruence classes. Thus $(\mathbb Z,+)/n\mathbb Z=(\mathbb Z/n,+)$ as the same group on the same underlying set. This includes $n=0$ and $n=1$. (For every $n\in\mathbb N$, the congruence-class group $(\mathbb Z/n,+)$ is the quotient group $(\mathbb Z,+)/n\mathbb Z$).

Given: The objects and hypotheses in the example.

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 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).

Let $G$ and $H$ be groups. Their external direct product has underlying set $G\times H:=\{(g,h):g\in G,\ h\in H\}$ and componentwise operation $(g,h)(g',h') := (gg',hh').$ The fact that this operation makes $G\times H$ a group, with the indicated identity and inverses, is proved in thm-external-direct-product-is-a-group. Until that result is used, this definition introduces only the set and its componentwise binary operation. (The external direct product $G\times H$ with componentwise multiplication).

For groups $G$ and $H$, the componentwise operation of def-external-direct-product-of-groups makes $G\times H$ a group. Its identity is $(e_G,e_H)$, and $(g,h)^{-1}=(g^{-1},h^{-1}).$ Moreover the coordinate maps $\pi_G(g,h)=g$ and $\pi_H(g,h)=h$ are group homomorphisms. ($G\times H$ is a group with identity $(e_G,e_H)$, coordinatewise inverses, and homomorphic coordinate projections).

Let $G$ be a group. For $g,h\in G$, their commutator is $[g,h]:=ghg^{-1}h^{-1}.$ This convention is fixed throughout; some sources use its inverse. By the inverse laws of lem-group-inverse-laws, one has $[g,h]^{-1}=hgh^{-1}g^{-1}=[h,g]$. The commutator subgroup, or derived subgroup, is the subgroup generated by all commutators: $[G,G]:=\langle\{[g,h]:g,h\in G\}\rangle.$ The generated subgroup notation is that of def-generated-subgroup. (Commutators $[g,h]=ghg^{-1}h^{-1}$ and the commutator subgroup $[G,G]$).

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).

For every $n\in\mathbb N$, view $n$ as its canonical nonnegative integer and put $n\mathbb Z:=\{nk:k\in\mathbb Z\}$. Then the left cosets of $n\mathbb Z$ in $(\mathbb Z,+)$ are exactly the congruence classes modulo $n$, and coset addition is the published addition of congruence classes. Thus $(\mathbb Z,+)/n\mathbb Z=(\mathbb Z/n,+)$ as the same group on the same underlying set. This includes $n=0$ and $n=1$. (For every $n\in\mathbb N$, the congruence-class group $(\mathbb Z/n,+)$ is the quotient group $(\mathbb Z,+)/n\mathbb Z$).

Given: The objects and hypotheses in the example.

Let $G=\langle X\mid R\rangle$ and $H=\langle Y\mid S\rangle$ with disjoint generators, and let $f,h$ embed $K$. If $T$ generates $K$ and words $u_t(X),v_t(Y)$ represent $f(t),h(t)$, then $G\ast_KH\cong\langle X\sqcup Y\mid R\cup S\cup\{u_t v_t^{-1}:t\in T\}\rangle.$ (A free product with amalgamation has the factor presentations plus the amalgamating relations).

The canonical maps $G\to G\ast_KH$ and $H\to G\ast_KH$ are injective. (The factor maps into a free product with amalgamation are injective).

Every class in $W(X)/{\sim}$ contains exactly one reduced word. (Every class in $W(X)/{\sim}$ contains exactly one reduced word).

For every set $X$, the group $F_{\mathrm{word}}(X)=W(X)/{\sim}$ together with $i_{\mathrm{word}}(x)=[x]$ is a free group on $X$ in the sense of def-free-group. (The word-quotient group $W(X)/{\sim}$ satisfies the universal property of the free group on $X$).

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 $G$ be a group and $R\subseteq G$. Then $\langle\!\langle R\rangle\!\rangle_G=\left\{g_1r_1^{\varepsilon_1}g_1^{-1}\cdots g_nr_n^{\varepsilon_n}g_n^{-1}:n\in\mathbb N,\ g_i\in G,\ r_i\in R,\ \varepsilon_i\in\{1,-1\}\right\}.$ For $n=0$ the displayed product is the identity. Replacing every conjugator $g_i$ by $g_i^{-1}$ gives the equivalent convention $g_i^{-1}r_i^{\varepsilon_i}g_i$. (The normal closure of $R$ is the set of finite products of conjugates of elements of $R$ and their inverses).

Given: The objects and hypotheses in the example.

Let $G=\langle X\mid R\rangle$ and $H=\langle Y\mid S\rangle$ with disjoint generators, and let $f,h$ embed $K$. If $T$ generates $K$ and words $u_t(X),v_t(Y)$ represent $f(t),h(t)$, then $G\ast_KH\cong\langle X\sqcup Y\mid R\cup S\cup\{u_t v_t^{-1}:t\in T\}\rangle.$ (A free product with amalgamation has the factor presentations plus the amalgamating relations).

The canonical maps $G\to G\ast_KH$ and $H\to G\ast_KH$ are injective. (The factor maps into a free product with amalgamation are injective).

Inside $G\ast_KH$, the images of $G$ and $H$ intersect exactly in their common image of $K$. (The two factor images intersect exactly in the amalgamated subgroup).

Every subgroup $H$ of a cyclic group $G=\langle g\rangle$ is cyclic. If $H\ne\{e\}$, then the least positive integer $d$ for which $g^d\in H$ satisfies $H=\langle g^d\rangle$. (Every subgroup of a cyclic group is cyclic; the least positive exponent in a nontrivial subgroup supplies a generator).

Given: The objects and hypotheses in the statement.

For a pushout of $f:K\to G$ and $h:K\to H$, $i_H(h(\ker f))=\{e\},\qquad i_G(f(\ker h))=\{e\}.$ Hence canonical factor maps in an arbitrary group pushout need not be injective. No equality with their full kernels is asserted. (The kernels of the amalgamating maps are killed in the opposite canonical maps to a group pushout).

For homomorphisms $f:K\to G$ and $h:K\to H$, let $N$ be the normal closure in $G\ast H$ of $\{j_G(f(k))j_H(h(k))^{-1}:k\in K\}.$ Then $(G\ast H)/N$, with the induced factor maps $j_G$ and $j_H$, is a pushout of $f$ and $h$. (A group pushout is the quotient of a free product by the amalgamating relations).

For every $n\in\mathbb N$, view $n$ as its canonical nonnegative integer and put $n\mathbb Z:=\{nk:k\in\mathbb Z\}$. Then the left cosets of $n\mathbb Z$ in $(\mathbb Z,+)$ are exactly the congruence classes modulo $n$, and coset addition is the published addition of congruence classes. Thus $(\mathbb Z,+)/n\mathbb Z=(\mathbb Z/n,+)$ as the same group on the same underlying set. This includes $n=0$ and $n=1$. (For every $n\in\mathbb N$, the congruence-class group $(\mathbb Z/n,+)$ is the quotient group $(\mathbb Z,+)/n\mathbb Z$).

Given: The objects and hypotheses in the example.

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).

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 $\operatorname{Aut}(G)$).

The inverse of a bijective group homomorphism is a group homomorphism. If $f:G\to H$ is a bijective group homomorphism, then its set-theoretic inverse $f^{-1}:H\to G$ is a group homomorphism. (The inverse of a bijective group homomorphism is a group homomorphism).

For every $n\in\mathbb N$, view $n$ as its canonical nonnegative integer and put $n\mathbb Z:=\{nk:k\in\mathbb Z\}$. Then the left cosets of $n\mathbb Z$ in $(\mathbb Z,+)$ are exactly the congruence classes modulo $n$, and coset addition is the published addition of congruence classes. Thus $(\mathbb Z,+)/n\mathbb Z=(\mathbb Z/n,+)$ as the same group on the same underlying set. This includes $n=0$ and $n=1$. (For every $n\in\mathbb N$, the congruence-class group $(\mathbb Z/n,+)$ is the quotient group $(\mathbb Z,+)/n\mathbb Z$).