free product of the same group

Question

Let $G$ and $H$ be two groups. a reduced word is an element of the form $w=g_1h_1g_2h_2...$ after removing an instance of the identity element (of either $G$ or $H$) and replacing a pair of the form $g_1g_2$ by its product in $G$, or a pair $h_1h_2$ by its product in $H$. Denote $G*H$ the set of reduced words $w$ on which we define a group law by concatenation of words and we call it the free product of $G$ and $H$.

when $G=H$ a word in $G*G$ is an element $w=g_1^1g_2^2g_3^1g_4^2...$ where $g_i^j$ is the element $g_i$ from the $j$th copy, $j=1,2$. Why we are not allowed to replace $g_1^1g_1^2g_2^1g_2^2...$ by the element $g$ who is the product $g_1^1g_2^2g_3^1g_4^2...$ since product $g_k^1g_l^2$ is well defined being in the same group $G$ ?

Answer 1

The free product is actually a coproduct in the category of groups, so it must satisfy a universal property. Indeed, $G \ast H$ is defined so that if there is any group $X$ along with homomorphisms $G \to X$ and $H \to X$, then there exist a unique homomorphism $\theta : G \ast H \to X$. If we were allowed to reduce all words in $G \ast G$, then the universal property would be broken (as an exercise, try to find a counter-example).

Answer 2

The short answer is that you're not allowed to do it because that's not the definition of the free product!

But let me try to explain: the free product of $G$ and $H$ is an "external" product of $G$ and $H$. It just doesn't depend on whether $G$ and $H$ are subgroups of some common group. You might as well ask why in the Cartesian product $\mathbb{Z} \times \mathbb{Z}$ you can't identify the element $1$ in the first factor with the element $1$ in the second factor. That's not how the Cartesian product works: the element $(1,0)$ is distinct from the element $(0,1)$.

What you're proposing would lead to a similarly incorrect description of the free product. For instance, according to your proposal $\mathbb{Z} * \mathbb{Z}$ would be an abelian group (indeed it would just be $\mathbb{Z}$), and not what it actually is: the free group on two generators.