So, I need to show that a finitely generated group isomorphic to the free product of two copies of itself (obviously thinking every factor as being generated by diferent letters) must be trivial.
I must clarify that is just a conjecture I made to complete a proof of another thing, but I think it is true. I have tried taking minimal sets of generators but didn't get too far.
Any help would be awesome.
Yes, it's true, and follows from Grushko's Theorem: $$\mathrm{rank} (A\ast B) = \mathrm{rank}A + \mathrm{rank}B.$$ (This is proved, for example, in Magnus, Karrass and Solitar, or on Wikipedia. The idea of the proof is that a minimal generating set for $A\ast B$ can be transformed by Nielsen transformations into one where a subset of generators belongs to $A$ and the rest belong to $B$.)
Thus, if $G$ has finite rank $n$, then $G\ast G$ has rank $2n$, so $n = 2n$ implies $n=0$, or $G=1$.