Why is the free product of a group with itself not isomorphic to itself? I've been recently introduced to the definition of free product of groups, and I'm struggling to identify the details of its precise definition. In particular, it seems to me from the definition I've seen that a free product of a group $G$ with itself, $G * G$, would be isomorphic to $G$ again. But I've seen that we speak of such things as $F_n$, the free product of $\mathbb{Z}$ with itself $n$ times, so obviously I must be wrong.
Below I'll write the definition of free product as I've understood it, trying to be as precise as possible and as formal as is pragmatic, and then show how (I think) that definition leads to the aforementioned result. I'd like to know what part of the definition I got wrong.
Definition of free product
Let $(G, \cdot_G)$ and $(H, \cdot_H)$ be groups. The free product of $(G, \cdot_G)$ and $(H, \cdot_H)$, denoted $(G, \cdot_G) * (H, \cdot_H)$, is the group $(G * H, \cdot_{G * H})$, where $G * H$ is the set of reduced words on $G$ and $H$, and $\cdot_{G * H}$ is the operation given by concatenation followed by reduction. Precise definitions of each are given below.

*

*The set of reduced words on $G$ and $H$, denoted $G * H$, is defined as $r_{G,H}[W(G \cup H)]$, where $W(\cdot)$ denotes the set of words on a set and $r_{G,H}$ is the function that maps each word to its reduced form. That is, it is the image of the set of words on $G \cup H$ under the reduction map $r_{G, H}$.

*

*Let $S$ be a set. The set of words on $S$, denoted $W(S)$, is defined as the set of finite but nonempty (ordered) tuples in elements of $S$. That is:
$$W(S) := \{ (x_1, \ldots, x_n) \mid n \in \mathbb{Z}_{>0},\quad \forall i \in \{1, \ldots, n\}\ x_i \in S \}$$
(The words are usually denoted as a formal product $x_1x_2\cdots x_n$, but I'm going to avoid this notation here to be clear how they are formally defined and so as to not be tempted to confuse the formal product with the product in $G$ in the next section.) Thus, in particular: $$W(G \cup H) = \{ (x_1, \ldots, x_n) \mid n \in \mathbb{Z}_{> 0},\quad \forall i \in \{1, \ldots, n\}\ (x_i \in G \lor x_i \in H) \}$$


*The reduction map $r_{G,H}$ is the function $r_{G,H} \colon W(G \cup H) \to W(G \cup H)$ that maps each word to its reduced form. The reduced form of a word is the word that we obtain by repeatedly applying any of the following operations until we can no longer apply any of them.

*

*Omit the identity of $G$, $1_G$, from a word of lenght greater than 1. That is, transform a word of the form $(x_1, \ldots, x_{i - 1}, 1_G, x_{i + 1}, \ldots, x_n)$ (with $n > 1$) into the word $(x_1, \ldots, x_{i - 1}, x_{i + 1}, \ldots, x_n)$.

*Omit the identity of $H$, $1_H$, from a word of lenght greater than 1. That is, transform a word of the form $(x_1, \ldots, x_{i - 1}, 1_H, x_{i + 1}, \ldots, x_n)$ (with $n > 1$) into the word $(x_1, \ldots, x_{i - 1}, x_{i + 1}, \ldots, x_n)$.

*Replace two consecutive elements of $G$, $g_1$ and $g_2$, with their product $g_1 \cdot_G g_2$. That is, transform a word of the form $(x_1, \ldots, x_{i - 1}, g_1, g_2, x_{i + 2}, \ldots, x_n)$, where $g_1, g_2 \in G$, into the word $(x_1, \ldots, x_{i - 1}, g_1 \cdot_G g_2, x_{i + 2}, \ldots, x_n)$.

*Replace two consecutive elements of $H$, $h_1$ and $h_2$, with their product $h_1 \cdot_H h_2$. That is, transform a word of the form $(x_1, \ldots, x_{i - 1}, h_1, h_2, x_{i + 2}, \ldots, x_n)$, where $h_1, h_2 \in H$, into the word $(x_1, \ldots, x_{i - 1}, h_1 \cdot_H h_2, x_{i + 2}, \ldots, x_n)$.

(I omit the proof that there is a unique reduced form, etc.)




*The operation $\cdot_{G * H}$ is defined by concatenating and then reducing. That is, if $(x_1, \ldots, x_n)$ and $(y_1, \ldots, y_m)$ are elements of $G * H$, then: $$(x_1, \ldots, x_n) \cdot_{G*H} (y_1, \ldots, y_m) := r_{G,H}\big((x_1, \ldots, x_n, y_1, \ldots, y_n)\big)$$
Proof of $G * G \cong G$ from the above definition
Let $(G, \cdot)$ be a group, and consider the free product of $G$ with itself, $(G * G, \cdot_{G * G})$. By the above definition, the set $G * G$ is the set $r_{G,G}[W(G \cup G)] = r_{G,G}[W(G)]$, so it is the set of reduced words in $G$. Let us consider an arbitrary element of $W(G)$, which will be of the form $(g_1, \ldots, g_n)$ with $g_i \in G$. Its reduced form will just be the 1-tuple $(g,)$ where $g = g_1 \cdot g_2  \cdots g_n$. Indeed, since all the elements of the word are in $G$, we can apply operation 3 above (or operation 4, which is the same in this case since $H$ "is" $G$) repeatedly until we're left with a word of length 1:
$$(g_1, g_2, g_3 \ldots, g_n) \rightsquigarrow (g_1g_2, g_3 \ldots, g_n) \rightsquigarrow \cdots \rightsquigarrow (g_1 \cdots g_n,)$$
Since the set of all finite products of elements in $G$ is just $G$ again, we thus have that $G*G = r[W(G)] = \{(g,) \mid g \in G\}$. Moreover, $G$ is isomorphic to $G * G$ via the natural map $\phi \colon G \to G*G$ defined by $g \mapsto (g,)$. Indeed, if $g_1, g_2 \in G$, then by what we have said about reduced forms: $$\phi(g_1 \cdot g_2) = (g_1 \cdot g_2,) = r_{G,G}\big((g_1, g_2)\big) \doteq (g_1,) \cdot_{G*G} (g_2,) = \phi(g_1) \cdot_{G*G} \phi(g_2)$$ so $\phi$ is a homomorphism (and $\phi$ is clearly bijective).
Guess
My guess is that the union in the above definition should be disjoint, i.e. that $G*H$ should be defined as $r[W(G \sqcup H)]$, adjusting the rest of the definition accordingly?
 A: As others have pointed out in the comments, the words should be over the disjoint union of $G$ and $H$. For the sake of completeness, I'm posting an answer to write down what the correct definition would be.
Definition of free product
Let $(G, \cdot_G)$ and $(H, \cdot_H)$ be groups. The free product of $(G, \cdot_G)$ and $(H, \cdot_H)$, denoted $(G, \cdot_G) * (H, \cdot_H)$, is the group $(G * H, \cdot_{G * H})$, where $G * H$ is the set of reduced words on $G$ and $H$, and $\cdot_{G * H}$ is the operation given by concatenation followed by reduction. Precise definitions of each are given below.

*

*The set of reduced words on $G$ and $H$, denoted $G * H$, is defined as $r_{G,H}[W(G \sqcup H)]$, where $W(\cdot)$ denotes the set of words on a set and $r_{G,H}$ is the function that maps each word to its reduced form. That is, it is the image of the set of words on $G \cup H$ under the reduction map $r_{G, H}$.

*

*Let $S$ be a set. The set of words on $S$, denoted $W(S)$, is defined as the set of finite (ordered) tuples in elements of $S$. That is:
$$W(S) := \{ (x_1, \ldots, x_n) \mid n \in \mathbb{Z}_{\ge 0},\quad \forall i \in \{1, \ldots, n\}\ x_i \in S \}$$
Thus, in particular:
$$W(G \sqcup H) = \{ (x_1, \ldots, x_n) \mid n \in \mathbb{Z}_{\ge 0},\quad \forall i \in \{1, \ldots, n\}\quad (\exists g \in G\ x_i = (g, 1))\ \veebar\ (\exists h \in H\ x_i = (h, 2))\}$$


*The reduction map $r_{G,H}$ is the function $r_{G,H} \colon W(G \sqcup H) \to W(G \sqcup H)$ that maps each word to its reduced form. The reduced form of a word is the word that we obtain by repeatedly applying any of the following operations until we can no longer apply any of them.

*

*Omit a letter that is the identity in $G$ or in $H$ from a word. That is, transform a word of the form $(x_1, \ldots, x_{i - 1}, (1_G, 1), x_{i + 1}, \ldots, x_n)$ or $(x_1, \ldots, x_{i - 1}, (1_H, 2), x_{i + 1}, \ldots, x_n)$ into the word $(x_1, \ldots, x_{i - 1}, x_{i + 1}, \ldots, x_n)$.

*Replace two consecutive elements of the same group (either $G$ or $H$) with their product. That is, transform a word of the form $(x_1, \ldots, x_{i - 1}, (y_1, \iota), (y_2, \iota), x_{i + 2}, \ldots, x_n)$, where $y_1, y_2$ are both in $G$ or both in $H$ (since $\iota = \iota$) into the word $(x_1, \ldots, x_{i - 1}, (y_1 \cdot y_2, \iota), x_{i + 2}, \ldots, x_n)$, where $\cdot$ denotes correspondingly either $\cdot_G$ or $\cdot_H$.

(I omit the proof that there is a unique reduced form, etc.)




*The operation $\cdot_{G * H}$ is defined by concatenating and then reducing. That is, if $(x_1, \ldots, x_n)$ and $(y_1, \ldots, y_m)$ are elements of $G * H$, then: $$(x_1, \ldots, x_n) \cdot_{G*H} (y_1, \ldots, y_m) := r_{G,H}\big((x_1, \ldots, x_n, y_1, \ldots, y_n)\big)$$
Proof sketch that it is a group

*

*Closure under the operation. The product of two words is by definition the reduction of their concatenation, so it is also a word in reduced form.


*Associativity. The reduction map $r$ is "sub-idempotent", i.e. applying $r$ to any sub-word of a word and then applying it to the entire word is the same as just applying it to the entire word (details omitted). Thus:
$$((x_1, \ldots, x_r) \cdot_{G*H} (y_1, \ldots, y_s)) \cdot_{G*H} (z_1, \ldots, z_t) = r(r((x_1, \ldots, x_r) \mid (y_1, \ldots, y_s)) \mid (z_1, \ldots, z_t)) = r((x_1, \ldots, x_r) \mid (y_1, \ldots, y_s) \mid (z_1, \ldots, z_t)) = r((x_1, \ldots, x_r) \mid r((y_1, \ldots, y_s) \mid (z_1, \ldots, z_t))) = (x_1, \ldots, x_r) \cdot_{G*H} ((y_1, \ldots, y_s) \cdot_{G*H} (z_1, \ldots, z_t))$$
where $\mid$ denotes concatenation.


*Identity. The empty word $()$ is the identity, since concatenating with $()$ does nothing, and then reducing doesn't do anything either because the word was already in reduced form.


*Inverses. The inverse of $((x_1, \iota_1), \ldots, (x_n, \iota_n))$ is $((x_n^{-1}, \iota_n), \ldots, (x_1^{-1}, \iota_1))$.
