Postby **owlpride** » Fri Jan 20, 2012 12:47 am

Notice that G is a cyclic group generated by i (isomorphic to Z/4Z if you prefer). Hence a homomorphism f: G -> G is uniquely determined by f(i). There are 4 of these:

f(i) = 1 (-> f(z) = z^0)

f(i) = i (-> f(z) = z^1)

f(i) = -1 (-> f(z) = z^2)

f(i) = -i (-> f(z) = z^3)

How do you match up the homomorphism induced by f(i) with a polynomial f(z) = z^k? That's simple. You know that z^k describes a homomorphism for all k (that's simply using the group structure). Since G is cyclic, you know that two homomorphisms agree if they agree on a generator of G. So you only need to check the value of the polynomials on i.

A similar procedure would work for a more complicated abelian group. The structure theorem for finitely generated abelian groups promises you a finite set of independent generators of G. Find such a generating set explicitly. Each of these generators can be mapped to any element of the group whose order divides the order of the generator; and each such map extends uniquely to a map on the entire group.

In this context it's worthwhile to mention the universal property of free abelian groups: If F is free abelian (e.g. finitely generated and torsion-free) with basis v1, v2, ... and G is any other abelian group, then any map of sets {v1, v2, ...} -> G can be uniquely extended to a group homomorphism F -> G.

There's an analogue for non-abelian groups: If F is free on the generators v1, v2... and G is any other group (abelian or not), then any set map {v1, v2, ... } -> G can be uniquely extended to a homomorphism F -> G.