Products, coproducts, and free objects in groups and abelian groups

Last post we discussed products, coproducts, and free objects generally in category theory. This post we focus our attention on \catname{Grp} and \catname{Ab}.

\begin{definition} For a family of groups $\{G_i \mid i \in I\}$, define a binary operation on the Cartesian product as follows. If $f, g \in \prod_{i \in I} G_i$, $fg: I \mapsto \bigcup_{i \in I} G_i$ is the function given by $i \mapsto f(i) g(i)$. $\prod_{i \in I} G_i$ is called the direct product of the family of groups. \end{definition} \begin{definition} The weak direct product of groups $\{G_i \mid i \in I\}$ denoted $\prod_{i \in I}^w G_i$ is the set of all $f \in \prod_{i \in I} G_i$ such that $f(i) = e_i$ for all but finitely many $i \in I$. If all the groups $G_i$ are Abelian, $\prod_{i \in I}^w G_i$ is usually called the direct sum and denoted $\bigoplus_{i \in I} G_i$. \end{definition}

It is trivial to prove that $\prod_{i \in I} G_i$ is a product in \catname{Grp} and \catname{Ab}. It is also easy to see that the direct sum of Abelian groups is a coproduct in \catname{Grp}.

These definitions are external direct products or external direct sums, but sometimes a group has the direct product or direct sum structure within itself, and we may call it an internal weak direct product or internal direct sum in that case.

\begin{theorem} Let $\{N_i \mid i \in I\}$ be a family of normal subgroups of $G$ such that \begin{enumerate} \item $G = \left\langle \bigcup_{i \in I} N_i \right\rangle$ \item $\forall k \in I, N_k \cap \left\langle \bigcup_{i \neq k} N_i \right\rangle = \langle e \rangle$ \end{enumerate} Then $G \cong \prod_{i \in I}^w N_i$. \end{theorem} \begin{proof} If $(a_i)_{i \in I} \in \prod_{i \in I}^w N_i$, then $a_i = e_i$ except for finitely many $i \in I$. Let $I_0 = \{ i \in I \mid a_i \neq e_i\}$. $\prod_{i \in I_0} a_i$ is a well-defined element of $G$, since for $a \in N_i, b \in N_j, i \neq j, ab = ba$. Consequently, $\phi: \prod_{i \in I}^w N_i \to G, a \mapsto \prod_{i \in I_0} a_i$ and $e \mapsto e$ is a homomorphism such that $\phi \iota_i (a_i) = a_i$ for $a_i \in N_i$. Since $G$ is generated by the $N_i$'s, every element $a \in G$ is a finite product of elements from various $N_i$. $a$ can be expressed as $\prod_{i \in I_0} a_i$. Hence $\prod_{i \in I_0} \iota_i (a_i) \in \prod_{i \in I}^w N_i$ and $\phi \left(\prod_{i \in I_0} \iota_i (a_i) \right) = \prod_{i \in I_0} a_i = a$. Therefore $\phi$ is an epimorphism. Suppose that $\phi(a) = \prod_{i \in I_0} a_i = e \in G$. Assume for convenience that $I_0 = \{1,2,\cdots,n\}$. Then $\prod_{i \in I_0} a_i = a_1 a_2 \cdots a_n = e$. Hence $a_1^{-1} = a_2 \cdots a_n \in N_1 \cap \left\langle \bigcup_{i=2}^n N_i \right\rangle = \langle e \rangle$ and therefore $a_1 = e$. Repetition of this argument shows $a_i = e$ for all $i \in I$. Hence $\phi$ is a monomorphism. \end{proof} In light of this theorem, we have the following definition: \begin{definition} Let $\{N_i \mid i \in I\}$ be a family of normal subgroups of $G$ such that $G = \left\langle \bigcup_{i \in I} N_i \right\rangle$ and $\forall k \in I, N_k \cap \left\langle \bigcup_{i \neq k} N_i \right\rangle = \langle e \rangle$. Then $G$ is said to be the internal weak direct product of $\{N_i \mid i \in I\}$. \end{definition} We shall now construct a group $F$ that is free on the set $X$. Let $X = \emptyset$, then $F$ is the trivial group. If $X \neq \emptyset$, let $X^{-1}$ be a set disjoint from $X$ such that $|X| = |X^{-1}|$. Choose a bijection $X \to X^{-1}$ and denote the image of $x$ by $x^{-1}$. Choose a singleton $\{1\}$ that is disjoint from $X \cup X^{-1}$. A word on $X$ is a sequence $(a_1, a_2, \cdots)$ with $a_ i \in X \cup X^{-1} \cup \{1\}$ such that for some $n \in \mathbb{N}$, $a_k = 1$ for all $k \geq n$. The constant sequence is called the empty word and is denoted $1$. A word $(a_1, a_2, \cdots)$ on $X$ is said to be reduced iff \begin{enumerate} \item $\forall x \in X$, $x$ and $x^{-1}$ are not adjacent \item $a_k = 1$ implies $a_i = 1$ for all $i \geq k$. \end{enumerate}

Every nonempty reduced word is of the form $(x_1^{\lambda_1}, x_2^{\lambda_2}, \cdots, x_n^{\lambda_n}, 1, 1, \cdots)$ where $n \in \mathbb{N}$, $x_i \in X$, $\lambda_i \in \{-1,1\}$. Hereafter, this word is denoted by $x_1^{\lambda_1} x_2^{\lambda_2} \cdots x_n^{\lambda_n}$. Two reduced words $x_1^{\lambda_1} x_2^{\lambda_2} \cdots x_m^{\lambda_m}$ and $y_1^{\delta_1} y_2^{\delta_2} \cdots y_n^{\delta_n}$ are equal iff both are 1 or $m = n$ and $x_i = y_i, \lambda_i = \delta_i$ for each $i \in \{1,2,\cdots,n\}$. Consequently the map from $X$ into the set $F(X)$ of all reduced words on $X$ given by $x \mapsto x^1 = x$ is injective. Identify $X$ with its image and consider it to be a subset of $F(X)$. Define a binary operation on the set $F = F(X)$ of reduced words on $X$ by juxtaposition and cancellations of adjacent terms.

\begin{theorem} If $X$ is a nonempty set, $F = F(X)$ is the set of all reduced words on $X$, then $F$ is a group and $F = \langle X \rangle$. \end{theorem} \begin{proof} $1$ is an identity element and $x_1^{\delta_1} x_2^{\delta_2} \cdots x_n^{\delta_n}$ has inverse $x_n^{-\delta_n} \cdots x_1^{-\delta_1}$. $\forall x \in X, \delta \in \{-1,1\}$, let $|x^\delta|: F \to F$ be given by $1 \mapsto x^\delta$ and \[ x_1^{\delta_1} x_2^{\delta_2} \cdots x_n^{\delta_n} \mapsto \begin{cases} x^\delta x_1^{\delta_1} \cdots x_n^{\delta_n} & x^\delta \neq x_1^{-\delta_1} \\ x_2^{\delta_2} \cdots x_n^{\delta_n} & x^\delta = x_1^{-\delta_1} \end{cases} \] Let $A(F)$ be the group of permutations on $F$ and $F_0$ the subgroup generated by $\{|x| \mid x \in X\}$. The map $\phi: F \to F_0$ given by $1 \mapsto \text{id}_F$, $x_1^{\delta_1} x_2^{\delta_2} \cdots x_n^{\delta_n} \mapsto |x_1^{\delta_1}| \cdots |x_n^{\delta_n}|$ is a surjection such that $\phi(w_1 w_2) = \phi(w_1) \phi(w_2)$. Since $1 \mapsto x_1^{\delta_1} x_2^{\delta_2} x_n^{\delta_n}$ under the map $|x_1^{\delta_1}| \cdots |x_n^{\delta_n}|$, $\phi$ is injective. The fact that $F_0$ is a group implies that associativity holds in $F$ and that $\phi$ is an isomorphism of groups. \end{proof} Some facts about free groups: if $|X| \geq 2$, then the free group is nonabelian. Every element except $1$ has infinite order. Every subgroup of a free group is itself a free group on some set. \begin{theorem} Let $F$ be the free group on a set $X$ then $F$ is a free object on the set $X$ in \catname{Grp}. \end{theorem} \begin{proof} Let $G$ be a group and $f: X \to G$. Define $\bar{f}: F \to G$ to be $\bar{f}(1) = e$ and for $x_1^{\delta_1} x_2^{\delta_2} \cdots x_n^{\delta_n}$ a nonempty reduced word on $X$, $$ \bar{f}(x_1^{\delta_1} x_2^{\delta_2} \cdots x_n^{\delta_n}) = f(x_1)^{\delta_1} f(x_2)^{\delta_2} \cdots f(x_n)^{\delta_n}$$ $\bar{f}$ is a homomorphism such that $\bar{f} \circ \iota = f$. If $g: F \to G$ is any homomorphism such that $g \circ \iota = f$, then $$ g(x_1^{\delta_1} x_2^{\delta_2} \cdots x_n^{\delta_n}) = g(x_1)^{\delta_1} g(x_2)^{\delta_2} \cdots g(x_n)^{\delta_n} = \bar{f}(x_1^{\delta_1} x_2^{\delta_2} \cdots x_n^{\delta_n})$$ Thus $\bar{f}$ is unique. \end{proof} \begin{corollary} Every group $G$ is the homomorphic image of a free group. \end{corollary} \begin{proof} Let $X$ be a set of generators of $G$ and let $F$ be the free group on the set $X$. The inclusion map $X \to G$ induces a homomorphism $\bar{f}: F \to G$ such that $x \mapsto x$. Since $G = \langle X \rangle$, $\bar{f}$ is an epimorphism. \end{proof} A consequence is that any group $G$ is isomorphic to a quotient group $F / N$. $F$ is the free group on $X$ and $N$ is the kernel of the epimorphism $F \to G$. $F$ is determined to isomorphism by $X$ and $N$ is determined by any subset that generates it as a subgroup of $F$. If $w = x_1^{\delta_1} x_2^{\delta_2} \cdots x_n^{\delta_n} \in F$ is a generator of $N$, then under the epimorphism $F \to G$, $w \mapsto x_1^{\delta_1} \cdots x_n^{\delta_n} = e$. The equation $x_1^{\delta_1} x_2^{\delta_2} \cdots x_n^{\delta_n} = e$ in $G$ is called a relation on the generators $x_i$. A given group $G$ may be completely described by specifying a set $X$ of generators of $G$ and a suitable set $R$ of relations on these generators. This description is not unique since there are many possible choices of both $X$ and $R$. Conversely, suppose we are given a set $X$ and a set $Y$ of reduced words on the elements of $X$. Let $F$ be a free group on $X$ and $N$ the normal subgroup of $F$ generated by $Y$ (intersection of all normal subgroups of $F$ containing $Y$). Let $G = F/N$ and identify $X$ with its image in $F/N$ under the map $X \hookrightarrow F \twoheadrightarrow F/N$. Then $G$ is a group generated by $X$ and all the relations $w = e$ are satisfied.