Our situation is following. Assume that we have free product $\star\_{i<n} G\_i$ each $G\_i$ finite group and assume that we have normal subgroup $K$ such that composition of canonical embedding and the quotinet mapping $G\_i\to \star\_{i<n}G\_i\to \star\_{i<n}G\_i/K$ is still embedding. Is there bigger normal subgroup $L>K$ such that $G\_i\to \star\_{i<n}G\_i\to \star\_{i<n}G\_i/L$ is still embedding and $\star\_{i<n}G\_i/L$ is finite?



embedding of finite groups into product
The answer is no, unless you restrict somehow the prime divisors of $|G\_i|$. Take $G\_i = \mathbb{Z}/p\_i$ where $p\_i$ is the $i$-th prime number. Take $K$ to be the trivial subgroup. Now, if you have such a subgroup $L$ then on the one hand $\*\_i G\_i/L$ is finite, and on the other hand, it contains an element of order $p\_i$ for every $i$, which is impossible.


