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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?

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1 Answer 1

up vote 7 down vote accepted

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.

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A slightly different way to put it (that doesn't generalize nicely) would be to consider the usual kernel $K$ of $f: G \to H$, the inclusion $i:K \to G$ and the map $j$ sending every element of $K$ to the neutral element in $G$. Then $fj = fi$, so $i = j$ and thus $K = \{e\}$. –  t.b. Apr 16 '12 at 19:16
Sorry, me again. I see how you can extend Arturo's proof to monoids or vector spaces - but how can it work on rings? –  MiMo Apr 19 '12 at 20:52
@Mimo: You use the free ring in one generator. In the category of rings with unity and unital morphisms, this is $\mathbb{Z}[x]$; in the category of rings, it's the subring of $\mathbb{Z}[x]$ generated by $x$ (i.e., "integer polynomials" in $x$ with no constant terms, where $kx$ represents $x+x+\cdots+x$, $k$ summands). –  Arturo Magidin Apr 19 '12 at 21:40
Thanks Arturo. It has been bugging me for a week, and right after I posted my comment I finally got it. I kept trying to build morphisms from Z to a ring, instead of thinking about all the possible combination of one element using the two operations (the "integer polynomials"). –  MiMo Apr 20 '12 at 7:16

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