TITLE: decidability of group homomorphism existence
QUESTION [7 upvotes]: Fix a finitely-presented group $G$ with distinguished non-identity element $g$.  For any finitely-presented group $H$ with element $h$, is it decidable whether there is a homomorphism $h: G \rightarrow H$ such that $h(g) = h\ ?$
If we know $G$ is cyclic, the question is undecidable by reduction from the Word Problem. But what if we don't know anything about $G$? What if we know $g$ has finite order in $G$?

REPLY [8 votes]: The problem is decidable if and only if there exists a homomorphism from $G$ to the infinite cyclic group $\mathbb Z$ taking $g$ to 1 (the generator of $\mathbb Z$). Clearly, if such a homomorphism exists, the answer to your question is "yes" for every $H,h$. Suppose that there is no such homomorphism.  Consider the signature (group operations, nullary operation) of pairs $(H,h)$ where $H$ is a group, $h\in H$. Let $X$ be the set of generators and $R$ be the set of defining relations of $G$, suppose $g$ is represented by a word $w$ in $X$. Then you are asking, whether a formula $\theta=\exists X (\& R \& (w=h))$ is true for the pair $(H,h)$. But the negation $\neg\theta$ is a Markov property. Indeed, there exists a pair, say, $({\mathbb Z},1)$ which satisfies $\neg\theta$ because there is no homomorphism $G\to \mathbb Z$ that takes $g$ to $1$, and there exists a pair, say, $(G,g)$ which cannot be embedded into any pair $(G',g)$ which satisfies $\neg\theta$. The proof that Markov properties for pairs $(H,h)$ are undecidable is the same as the proof of the Adian-Rabin theorem for groups .