Operations between subgroups in a free product of groups The free product is the coproduct in the category of groups. Take the construction of its elements using words as in https://en.wikipedia.org/wiki/Free_product.
Obviously, in a free product there can be no nontrivial result of operations between elements of different groups. For example, $g1h1$ could not be equal to $h2$, an element of $H$.
What in the universal mapping property of coproducts forbids this kind of operation in a free product ? The commutative diagrams (https://en.wikipedia.org/wiki/Coproduct) are defined from $H$ and $G$. Why would they not commute if additional "nontrivial" operations across "subgroups" (across images of $H$ and $G$ in $H * G$) are defined ?
Maybe an example could help me. If there is such an operation across subgroups, what would be violated in the definition of the coproduct ?
 A: There is nothing in the universal property which forbids this; this behavior is specific to groups, and in other categories of algebraic objects coproducts do allow this kind of "mixing." For example, the coproduct in the category of commutative rings is the tensor product, and "mixing" is possible here, e.g. the tensor product $\mathbb{F}_2 \otimes \mathbb{F}_3$ is zero so the image of every element of $\mathbb{F}_2$ is equal to the image of every element of $\mathbb{F}_3$. 
A: One way to see this is known as "van der Waerden's trick" (or a variant of it).  Namely, let $W$ be the set of reduced words in the definition, and consider the symmetric group $S(W)$.  Then for each element $g \in G$, we can define an element of $S(W)$ by: if $w \in W$ begins with $g' \in G$, then replace $g'$ with $(gg')$ (as a single initial symbol) and also remove it if it happens that $g' = g^{-1}$.  Otherwise, prepend $g$ to $w$.  (And of course, if $g = e$ then send $g$ to the identity permutation.)
It is straightforward (though involving several different cases to check, depending on whether the "removal" happens at each step or not) to prove that this induces a group homomorphism $G \to S(W)$.  We can also construct a group homomorphism $H \to S(W)$ in a similar way.  By the coproduct universal property, these two maps induce a unique group homomorphism $H * G \to S(W)$.
But now, if we have any $g \in G$, $h \in H$, then applying $h$ to the empty string results in $h$, and then applying $g$ results in the string $gh$.  Therefore, the image of $gh$ in $S(W)$ cannot be equal to the image of $h'$ in $S(W)$ which sends the empty string to just $h'$.  That implies that $gh \ne h'$ in $H*G$.  A very similar argument shows that in general, each reduced word in $W$, when converted to its corresponding element of $H*G$, has a distinct image in $S(W)$.
