Let $G_{1}$, $G_{2}$ be two groups given by their presentations. Then if $f$ is a map defined on the generators of $G_{1}$ we can prove that $f$ is a homomorphism by proving that the set of the relations of $G_{1}$ is mapped through $f$ to relations that hold in $G_{2}$. Is there a similar way to prove that $f$ is 1-1 by using the fact that we know the presentation of both groups?
2026-03-25 19:04:25.1774465465
Can you check if it's a group monomorphism from the group presentations?
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Nope. Showing that $f$ is 1-1 would amount to showing that for distinct $a$ and $b$ in $G_1$ that $f(a)$ and $f(b)$ are distinct in $G_2$. In general this is undecidable given just the presentation of $G_2$. See The word problem for groups.