Relations from quotient of free product This question arose from an exercise which asks you show that the fiber coproduct exists in the category of groups. I was eventually able to (mostly) solve the problem by “gluing” the images of elements under the two homomorphisms(i.e letting $\phi(h)\psi^{-1}(h)=1$ where $\phi : H \rightarrow G$ and $\psi : H \rightarrow G’$ are homomorphisms.)
I looked further into this concept (amalgamation) on the Wikipedia page and it says that in order to obtain the relation $\phi(h)\psi^{-1}(h)=1$ on the free product $G * G’$ you must quotient out by the smallest normal subgroup (i.e. the intersection of all normal subgroups) containing the words $\phi(h)\psi^{-1}(h)$. So my questions are: why isn’t it sufficient to simply adjust the definition of the binary operation so that whenever we have a product ($a_1a’_1…a’_n\phi(h)$)($\psi(h)b_1…b_kb’_k$) it is equivalent to $a_1a’_1…a_nb_1…b_kb’_k$ and how do we know that the normal subgroup isn’t “too big”(that is, it establishes an unwanted relation)?
I am familiar with relations established by quotients of free groups, however I don’t quite see why this works.
 A: Perhaps it helps to look more closely at the meaning of "the normal subgroup generated by" some subset.
Let $\Gamma$ be a group and $S \subset \Gamma$ a subset. Let $N$ be the smallest normal subgroup of $\Gamma$ that contain $S$, meaning the intersection of all normal subgroups of $\Gamma$ that contain $S$ (one should check that this intersection is indeed a normal subgroup that contains $S$).
One can write out a formula for a subset of $\Gamma$ that is a generating set for $N$, namely:
$$S^\Gamma = \{g^{-1} s g \mid s \in S, g \in \Gamma\}
$$
In all likelihood this is going to be an infinite set.
Let's prove that $N = \langle S^\Gamma \rangle$. First, using that $S \subset N$ together with normality of $N$ it follows that $S^\Gamma \subset N$, and therefore $\langle S^\Gamma \rangle < N$. In the other direction, the subset $S^\Gamma$ is clearly invariant under conjugation by elements of $\Gamma$: for each element $g^{-1} s g \in S^\Gamma$ and each $g' \in \Gamma$ we have
$$g'{}^{-1}(g^{-1}sg) g' = (gg')^{-1} s (gg') \in S^\Gamma
$$
It follows that the subgroup $\langle S^\Gamma \rangle$ is also invariant under conjugation by elements of $\Gamma$, which is just a long-winded way of saying that $\langle S^\Gamma \rangle$ is normal in $\Gamma$. Since $N$ is the intersection of all normal subgroups of $\Gamma$ that contain $S$, it follows that $N < \langle S^\Gamma \rangle$.
With that settled, let $\Gamma = G * G'$, let $S = \{\phi(h) \psi^{-1}(h) \mid h \in H\}$, and let $N$ be the subgroup of $\Gamma$ normally generated by $S$. Let's convince ourselves that $N$ does not "establish an unwanted relation". We have proved that $N$ is generated (in the ordinary sense) by $S^\Gamma$. Therefore every element of $N$ can be written as a product of elements of the form $g^{-1} s g$ for some $s \in S$, $g  \in G * G'$. Each such element, and any product of such elements, is clearly a "wanted" relation!
