In a finite field, is there ever a homomorphism from the additive group to the multiplicative group?

Question

I know that in infinite fields, such as $\mathbb{C}$, the mapping $e^x$ is a homomorphism from the additive group to the multiplicative group, and I was just wondering if in any finite field, there exists a (non trivial) homomorphism between the two.

Answer 1

This is impossible for finite fields.

Consider a finite field of order $q$; then the additive group also has order $q$. However, the multiplicative group has order $q - 1$ which does not share any common factors with $q$. Since the order of the image of an element $x$ under a homomorphism must divide the order of $x$ by Lagrange's theorem, it follows that any such homomorphism must be trivial.

Answer 2

Sometimes there is a homomorphism from the additive group of integers modulo N-1 and the multiplicative group of integers modulo N.

For example when N=5, given the additive group of four integers A = {0,1,2,3} and the finite field M = {0,1,2,3,4}, we can map each element a of A to a corresponding element m from the multiplicative group of M with m = 2^a mod 5.

With these groups, whenever z = x + y modulo (N-1),

it is also true that 2^z mod N = 2^x * 2^y modulo N.

(where addition, multiplication, exponentation, modulo, etc. are the common integer operations).

This is true whenever N is a Fermat prime. (This is related to the "invertable multiplication", also called "multiplication IDEA style", used in the IDEA cipher).

I suspect there may be other mappings that work for other prime values of N.

(The additive group of four integers A = {0,1,2,3} is not exactly the same as the finite field of four elements GF(4), but they both have 4 elements, so perhaps this is close to what you are looking for.)