Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

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.

share|improve this answer
4  
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
1  
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.