Composition of endomorphisms

1. Sep 25, 2005

Data

Hi. I haven't been here in a while, but I've just run across a question that I can neither find an easy counterexample for nor prove easily. Here it is:

Let G be a group, and $\alpha : G \rightarrow G$ and $\beta : G \rightarrow G$ be endomorphisms. Assume that $\alpha \circ \beta$ is an automorphism of $G$. Prove that $\alpha$ is injective and $\beta$ is surjective.

If G is finite the result is easy. I do not know how to prove it for infinite groups, nor have I been able to find a simple counterexample (it's easy to construct automorphisms that are compositions and for which the composing functions don't fill the injective/surjective criteria above, but I can't find a composition of homomorphisms that do it).

2. Sep 25, 2005

AKG

If b is not surjective, then a is not injective, since a(b(G)) = G and b(G) < G, so if a maps a subset of G to all of G, then the rest of G, G - b(G), must be sent to stuff that a has already sent b(G). If a is not injective, then there are g, h in G such that a(g) = a(h). If b is surjective, then b(G) = G and it contains g and h, so a*b will not be injective and hence not an automorphism. The two results above give us that a is injective iff b is surjective. I'm not sure that this is useful.

Now are you being asked by some text book to prove the result? If so, we shouldn't waste time looking for a counterexample, but is it possible that what you're being asked to prove is false?

Take G to be the product of infinite copies of Z2. Define a by:

a((x1, x2, ...)) = (x2, x3, ...)

Define b by:

b((x1, x2, ...)) = (0, x1, x2, ...)

a*b = I (identity) so it is clearly an automorphism. But a is not injective since it maps (0, x2, x3, ...) to the same place as (1, x2, x3, ...). b is not even surjective either, since it doesn't map anything to the 50% of G that starts with a 1 (using the figure "50%" rather loosely). And a and b are indeed endomorphisms, because whether you add numbers then push down (or pull forward), or whether you push down (or pull forward) and then add the numbers after doesn't matter. Unless there's some careless error in the above, it provides a counter-example to the given statement when |G| is not finite. Note that this agrees with the first result I gave that a is injective iff b is surjective. That result didn't really help, but I guess it gives me a little comfort because I haven't rigourously checked that my counterexample works.

3. Sep 26, 2005

Doodle Bob

I think it should be the other way around: $\alpha$ will necessarily be surjective, since $G=\alpha(\beta((G)))\subset \alpha(G)$ and $\beta$ will necessarily be one-to-one, since otherwise there would be two elements of G being sent to the same element under $\alpha \circ \beta$.

4. Sep 26, 2005

AKG

Yes, it is trivially true that a must be surjective and b must be injective. In the finite case, you can prove the not-so-trivial result that a must be injective and b must be surjective. In the infinite case, that becomes false.

5. Jan 9, 2006

Data

Hi again. I'm sorry that I abandoned this thread, first of all! Even though it's been so long, I figure that I had better post this for completeness' sake.

The problem turned out to be a misunderstanding. The source was writing compositions in the opposite way to that in which I (and everyone else I know) does, so it was actually just asking for the trivial result that $\alpha \cdot \beta$ is a bijection implies $\alpha$ is surjective and $\beta$ is injective.

And I don't see anything wrong with the counterexample posted above, either.

Thanks again!