I am trying to demonstrate that a monic group morphism is injective, but I am stuck - it seems it should be easy but I cannot get it.
Does anybody have a suggestion on how to proceed?
I am trying to demonstrate that a monic group morphism is injective, but I am stuck - it seems it should be easy but I cannot get it.
Does anybody have a suggestion on how to proceed?
This follows from the existence of a free group in one generator.
Let $f\colon G\to H$ be a monic group morphism, and assume that $f(x)=f(y)$. Let $h_x,h_y\colon \mathbf{Z}\to G$ be morphisms defined by $h_x(1) = x$ and $h_y(1) = y$, respectively (where $\mathbf{Z}$ is the additive group of integers, which is isomorphic to the free group in one generator). Then $f\circ h_x = f\circ h_y$ (since they agree on the generator of $\mathbf{Z}$). Since $f$ is monic, we conclude $h_x=h_y$, hence $x=y$. Thus, $f$ is injective on underlying sets.
The proof easily generalizes to any concrete category in which you have a free object on one generator (e.g., semigroups, monoids, lattices, rings, etc.).
For a nice example of a subcategory of $\mathbf{Groups}$ where monic does not imply injective take the category of divisible abelian groups, and show that $\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}$ is monic.