Amalgamated product Let $A$ and $B$ be two groups with a common subgroup $C$. Let $a_1,\dots,a_n\in A\setminus C$ and $b_1,\dots,b_n\in B\setminus C$. Is it true that $a_1b_1\cdots a_nb_n$ is not the identity in the amalgamated product $A*_CB$?
 A: Lyndon and Schupp, p186-187 is what you are after. Consider the free product with amalgamation $A\ast_{C_A=C_B}B$.
A sequence $g_1, \ldots, g_n$, $n\geq 0$, of elements of $A\ast B$ is called reduced if


*

*Each $g_i$ is in one of the factors $A$ or $B$.

*Successive $g_i, g_{i+1}$ come from different factors.

*If $n>1$, no $g_i$ is in $C$.

*If $n=1$, $g_1\neq 1$.


Clearly, what you have in front of you is (the concatenation of elements from) a reduced sequence. The relevant result is the following. Not that, due to a technicality, it isn't actually a normal form (for the group), and indeed even if $A$ and $B$ have soluble word problem it is not necessarily true that $A\ast_{C_A=C_B}B$ has soluble word problem (because you need the embeddings $C_A\leq A$ and $C_B\leq B$ to have certain properties also). Also, note that there is a subtlety in the theorem - before proving it we still need to distinguish between elements of $C_A$ and $C_B$.
Theorem 2.6. (Normal Form Theorem for Free Products with Amalgamation)
If $g_1, \ldots, g_n$ is a reduced sequence, $n\geq 1$, then the product $g_1\cdots g_n\neq1$ in $A\ast_{C_A=C_B}B$. In particular, $A$ and $B$ are embedded by the maps $a\mapsto a$, $b\mapsto b$.
Proof: I want to give the proof which is in Lyndon and Schupp, because I like it a lot. They introduce HNN-extensions before free products with amalgamation, and they utilise HNN-extensions in a beautiful way here. So:
Let $G=A\ast_{C_A=C_B}B$ be given by the relative presentation $\langle A, B; c=\phi(c), c\in C_A\rangle$ (the point of me saying this is that we now have a name for the isomorphism $\phi: C_A\rightarrow C_B$ we are pinning $A$ and $B$ across). Define the group $F^{\ast}=\langle A\ast B, t; t^{-1}ct=\phi(c), c\in C_A\rangle$. Then define the homomorphism $\Psi: G\rightarrow F^{\ast}$ as follows.
$$
\begin{cases}
\Psi(a)=t^{-1}at&\text{if }a\in A\\
\Psi(b)=b&\text{if }b\in B\\
\end{cases}
$$
This map is a homomorphism as all the defining relations of $G$ map to the identity. Apart from the case of $n=1$ and $g_1\in C_A\setminus\{1\}$, it is clear that every reduced sequence is mapped to a reduced sequence of the HNN-extension. However, if $g_1\in C_A\setminus\{1\}$ then $\Psi(g_1)=t^{-1}g_1t=\phi(g_1)\neq 1$. Therefore, the result follows from the normal form theorem for HNN-extensions.
A: Let $T$ be a transversal for $A/C$ and $S$ a transversal for $B/C$ minus an identity representative.
Let $M$ be the free monoid generated by the letters $\{t,s\}$ modulo the relations $s^2=s$ and $t^2=t$, and for $m\in M$ write $\Gamma(m)$ for the corresponding Cartesian product of $T$s and $S$s. For instance, $\Gamma(tst)= T\times S\times T$. Then the disjoint union $G=\bigsqcup_{m\in M}\Gamma(m)\times C$ can be endowed with the obvious multiplication operation, as elements of $C$ can be "slid past" representatives in $S$ and $T$ via the rule $cu=u'c'$ (with the right quantifiers, $\forall c\in C$, $\forall u\in T$ or $S$, $\exists u'\in T$ or $S$, $\exists c'\in C$).
One then checks that $G$ satisfies the universal property for the amalgamated product $A*_CB$. The claim is then clear because the coordinate vector $(a_1,b_1,\cdots)$ is not the identity $e\in C$.
