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.