46. Let G be the group of complex numbers 1, i, -1, -i under multiplication. Which of the following statements

are true about the homomorphisms of G into itself?

I. z ->z* defines one such homomorphism, where z* denotes the complex conjugate of z.

II. z-> z^2 defines one such homomorphism.

III. For every such homomorphism, there is an integer k such that the homomorphism has the form z ->z^k .

Hi, may I know how would we prove assertion (III)? Is it by considering cases? My group theory is not very strong, but I learnt that homomorphism must map 1 to 1, so that leaves us 3!=6 cases to consider?

Thanks a lot.

For (I), f(ab)=(ab)*=a*b*=f(a)f(b)

For (II), f(ab)=(ab)^2=(a^2)(b^2)=f(a)f(b).

## 0568 Q46

### Re: 0568 Q46

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

There's an analogue for non-abelian groups: If F is

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.### Re: 0568 Q46

Thanks a lot.

This tip is very useful!

This tip is very useful!

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.