There is something in the definition of the free product of two groups that annoys me, and it's this "word" thing:
If $G$ and $H$ are groups, a word in $G$ and $H$ is a product of the form
$$
s_1 s_2 \dots s_m,
$$where each $s_i$ is either an element of $G$ or an element of $H$.
So what is this "word" guy? Does it come out of the blue? Does it come from some sort of new operation that I can perform with the two sets $G$ and $H$ -in addition to the well-known ones of union, intersection, Cartesian product…?
Fortunatelly, I think there is nothing new under the sun of set operations: it's easy to realise that words can be identified with elements of some Cartesian product (see below):
$$
(s_1, s_2, \dots , s_m ) \ .
$$
And Cartesian product is a well-established set-theoretical operation.
So I tried to translate the rest of Wikipedia's definition
Such a word may be reduced using the following operations:
Remove an instance of the identity element (of either $G$ or $H$).
Replace a pair of the form $g_1g_2$ by its product in $G$, or a pair $h_1h_2$ by its product in $H$.Every reduced word is an alternating product of elements of $G$ and elements of $H$, e.g.
$$
g_1 h_1 g_2 h_2 \dots g_r h_r.
$$The free product $G ∗ H$ is the group whose elements are the reduced words in $G$ and $H$, under the operation of concatenation followed by reduction.
in an elementary set setting. First, consider the set of "unreduced" tuples of elements of $G$ and $H$
$$
U = G \sqcup H \sqcup (G\times G) \times (G\times H) \sqcup (H\times G) \sqcup (H\times H) \sqcup (G\times G \times G) \sqcup \dots
$$
More concisely:
EDIT:
I think the following formula may be less messier than the one I wrote previously:
$$
U = \bigsqcup_{r \geq 1} (S_1 \times \cdots \times S_r),
$$
where $S_i = G$ or $S_i = H$.
So, elements of $U$ are ordered tuples (unreduced ones)
$$
(s_1, s_2, \dots , s_m),
$$
where each $s_i$ is either an element of $G$ or an element of $H$.
The product of two unreduced tuples is defined by concatenation
$$
(s_1, \dots , s_m) \cdot (t_1, \dots , t_n) = (s_1, \dots , s_m, t_1 , \dots , t_n) \ .
$$
Now, consider the following equivalence relation in the set of unreduced tuples $U$:
$$
(s_1, s_2, \dots , s_{i-1}, 1, s_{i+1}, \dots , s_n) \sim (s_1, s_2, \dots, s_{i-1}, s_i, \dots , s_n) \ ,
$$
where $1$ is either the unit element of $G$ or the one of $H$. And
$$
(s_1, s_2, \dots , s_i,s_{i+1}, \dots , s_r) \sim (s_1, s_2, \dots , s_is_{i+1}, \dots , s_r )
$$
whenever two adjacent $s_i, s_{i+1} \in G$ or $s_i, s_{i+1} \in H$ at the same time.
If you want, you may call the equivalence class of a tuple under this equivalence relation a reduced tuple. So every reduced tuple is an alternating one,
$$
(g_1, h_1, \dots , g_r , h_r) \ ,
$$
with $g_i \in G$ and $h_i \in H$ for all $i = 1, \dots , r$.
Define the free product of $G$ and $H$ as the quotient:
$$
G*H = U/\sim \ .
$$
Finally, one verifies that concatenation is well-defined on unreduced tuples and gives $G*H$ a group structure.
After performing this elementary exercise I understand perfectly well why nobody defines the free product in this way, but I still wanted to ask:
- Is this correct?
- Is it written somewhere?
Best Answer
In answer to your question about whether this is written somewhere: the construction of free products in my book Introduction to Topological Manifolds proceeds very much along these lines. (I'm not the only one who's written it down, but I don't have other references handy.) I don't go through the set-theoretic description of the set of ordered tuples in as much detail as you do, but it can be described more simply as just the disjoint union of the sets $(G\amalg H)^n$ over all n. I like the concreteness of this construction as a way of giving one a handle on what the free product looks like. Once the group is constructed, it's a relatively easy matter to prove after the fact that it is uniquely determined up to isomorphism as the coproduct in the category of groups.