The following question is taken from Arrows, Structures and Functors the categorical imperative by Arbib and Manes
Definition for $\textit{Coproducts in}$ $\textbf{Mon}$ $\textit{and}$ $\textbf{Grp:}$ The one element monoid/group $\{e\}$ is clearly initial object - that is, the empty coproduct = in both categories. Let now $\{X_i\mid i\in I\}$ be a family of monoids with $I$ non-empty. LEt $A$ (for 'alphabet') be the disjoint union of the sets $X_i$. Let $W$ be the set $A^*$ of all words on the alphabet $A.$ A word $(a_1,\ldots,a_n)$ is $\textit{reduced}$ if for all $j<n$, $i_j\neq i_{j+1}$ and for all $j\leq n, x_j\neq e_j$ (where $a_j=(x_j,i_j)$ and $e_j$ is the unit of $X_{ij})$. If a word is not reduced it has a $\textit{reduction}$ obtained by continuing to apply the following rules:
(a) Multiply adjacent pairs with the same $i-$term:
$a_1, \ldots, a_j)=(x_j,i), a_{j+}=(x_{j+1},i),\ldots, a_n)\mapsto (a_1,\ldots,\hat{a}_{j}=(x_jx_{j+1},i),a_{j+2},\ldots,a_n$
(b) Delete identity terms:
$$(a_1,\ldots,a_j=(e_j,i_j),\ldots,a_n)\mapsto (a_1,\ldots,a_{j-1},a_{j+1},\ldots,a_n)$$ until the word is reduced.
Define a multiplication on $C$ by
$(a_1,\ldots,a_n)(b_1,\ldots b_m)=$reduction of $(a_1,\ldots a_n,b_1,\ldots,b_m)$.
Then $C$ is a monoid with unit $\wedge.$ Define injection homomorphisms
${in}_i:X_i\rightarrow C \quad x_i\mapsto$ reduction of $(x_i,i)$
These $\textit{are}$ monoid homomorphisms by the reduction rules. To check that $({in}_j:X_i\rightarrow C\mid i\in I)$ is indeed a coproduct for the monoids $X_i$ We must check that whenever $f_i:X\rightarrow D$ are monoid homomorphisms, the diagram
commutes for a unique monoid homomorphism $f.$ But for $f$ to be a homomorphism it clearly must satisfy $f(\wedge)=e,$ while, for $n>0, f(a_1,\ldots, a_n)=f_{i_1}(x_1)\ldots f_{i_n}(x_n)$ where $a_j=(x_j,i_j).$ But this determines $f$ as a function.
Exercise: Let $X$ be the two-element group $\{e,a\} (aa=e)$. Prove that $X+X$ is infinite.
Questions. I thought if I let $X_1=\{e,a\}$, $X_2=\{e,b\}$, then $X_1+X_2=:$ $ab, aab, abab, aabb, ababab,...$ Basically there would be an infinite number of string of words according to the definition of coproduct for monoids/groups up above. I am told that the coproduct for groups and monoids, is basically the free product. So if $X_1=X_2,$ I would just get strings of word of the form: $a,$ $aa,$ $aaa,$ $aa,\ldots a,\ldots.$ I am not sure how to go about doing the exercise since I am not sure what it means by showing the coproduct or direct sum to be infinite. If anyone can provide with with some direction and clarifications please.
Thank you in advance
Consider the element $ab \in X_1+X_2$, and the cyclic subgroup $\langle ab\rangle$ generated by $ab$. As you may notice $ab$ has no finite order, since $\smash[b]{\underbrace{ab\cdots ab\,}_\text{$n$ times}}\neq e$ for any $n \in \mathbb{N}$, by explicit construction of $X_1+X_2$.
Then $X_1+X_2$ contains the infinite cyclic subgroup $\langle ab\rangle\cong \mathbb{Z}$, and so $X_1+X_2$ is infinite.
Notice that $X_1+X_2=\mathbb{Z}/2\mathbb{Z} \ \ast \ \mathbb{Z}/2\mathbb{Z}$, called the infinite dihedral group.