Suppose $C\leq A,B$ are finitely generated groups of finite exponent. Can $A$ and $B$ be [amalgamated](https://en.wikipedia.org/wiki/Amalgamation_property) over $C$ in a group of finite exponent? What about if $A,B$ are periodic, can we find a periodic amalgam?


Note that if we assume that $A$ and $B$ are finite, then there is a finite amalgam by a classical construction of Neumann.


If we do not assume that the groups be finitely generated, then taking $C$ to be an infinite dimensional vector space over the field with $2$ elements, and $A,B$ to be appropriate expansions of $C$ by involutive automorphisms, we get an example where $A,B$ are of exponent 4 and every amalgam has an element of infinite order.


Thus, if this is true, the proof must use local finiteness, and if not, counterexamples must be (necessarily countably) infinite.


We could try to emulate the counterexample if we could find a f.g. group $C$ of finite exponent (or periodic) with two automorphisms of finite order whose product has infinite order (or generate jointly automorphisms of arbitrarily high order). But even then we would need to ensure that the group generated by $C$ together with each of these automorphisms is still finite exponent/periodic.


(I did not check carefully, but possibly, the conclusion is also true if we assume that $C$ is one of finite, finite index in one of $A,B$, or central in one of $A,B$, but I do not want to assume anything like that.)



Amalgamation of finitely generated finite exponent groups
It's not hard to construct counter-examples for large exponents.


Consider the following two automorphisms $\xi, \eta$ of the free group $F=F(x,y)$ of rank $2$, defined by
$$\xi(x)=y,~\xi(y)=x \text{ and } \eta(x)=x,~\eta(y)=yx.$$
The composition $\xi \circ \eta \in Aut(F)$ is the so-called Fibonacci automorphism $\varphi$, given by
$$\varphi(x)=y,~\varphi(y)=xy.$$


Now, for every $n \in \mathbb{N}$ the free Burnside group $B(2,n)$, of exponent $n$, is the quotient of $F$ by the verbal subgroup generated by all $n$-th powers. It follows that every automorphism of $F$ induces an automorphism of $B(2,n)$ (there is a natural homomorphism $Aut(F) \to Aut(B(2,n))$, $\alpha \mapsto \bar\alpha$).


Now, take $n$ be to sufficiently large and odd. Then it is known that the Fibonacci automorphism $\bar\varphi$ has infinite order in $Aut(B(2,n))$. This result was proved in
[*Cherepanov, E. A.*, [**Free semigroup in the group of automorphisms of the free Burnside group.**](https://doi.org/10.1081/AGB-200047435), Commun. Algebra 33, No. 2, 539-547 (2005). [ZBL1121.20028](https://zbmath.org/?q=an:1121.20028).[, and was improved (for smaller exponents) in
[*Pajlevanyan, Ashot S.*, [**The Fibonacci automorphism of free Burnside groups.**](https://doi.org/10.1051/ita/2011118), RAIRO, Theor. Inform. Appl. 45, No. 3, 301-309 (2011). [ZBL1227.20038](https://zbmath.org/?q=an:1227.20038).]


Evidently $\bar\xi$ has order at most $2$ and $\bar\eta$ has order dividing $n$ in $Aut(B(2,n))$, because the order of the image of $x$ divides $n$ (in fact, the orders of $\bar\xi$ and $\bar\eta$ will be exactly $2$ and $n$). Therefore, the semidirect products $G=B(2,n) \rtimes\_{\bar\xi} C\_2$ and $H=B(2,n)\rtimes\_{\bar\eta} C\_n$, where $C\_n$ is the cyclic group of order $n$, have exponents dividing $2n$ and $n^2$ respectively. It's easy to see that both $G$ and $H$ are generated by $2$ elements.


However, as the OP suggested in the post, since the composition $\bar \varphi$, of the automorphisms $\bar\xi$ and $\bar \eta$, has infinite order in $Aut(B(2,n))$, the groups $G$ and $H$ cannot be amalgamated over $B(2,n)$ in a periodic group.


