# Finite field structure

1. Sep 5, 2015

### Dragonfall

Take a prime order cyclic group. I want to take that as the additive group of a finite field. Since every finite field of the same order is isomorphic to one another, does the isomorphism define a multiplicative group structure on my cyclic group elements?

2. Sep 5, 2015

### andrewkirk

If every field of order n is isomorphic then there can only be one possible multiplicative structure on the prime cyclic additive group, and hence the isomorphism is just the identity. Is that what you mean?

3. Sep 5, 2015

### Dragonfall

Yes. But I forgot an additional criterion.

In the additive group, we can "multiply" by an integer n as nx= x+...+x, n times. Does the multiplication * defined by said isomorphism always satisfy

$$(ax)*(by) = (x+...+x)*(y+...+y) = (ab)(xy) = xy+...+xy$$

For positive integers a, b and group elements x, y?

4. Sep 5, 2015

### HallsofIvy

In other words, your choice to take the additive group of integers modulo p, a prime, as the addition pretty much forces you to take multiplication modulo p as the multiplication.

(Notice that the additive group of integers, modulo n, where n is not a prime, is a perfectly good group but the multiplication has zero-divisors so does not give you a field.)

5. Sep 5, 2015

### andrewkirk

'Multiplication' by repeated addition is not the same as multiplication of group elements. The additive group is a module over the ring of integers $\mathbb{Z}$. The repeated addition of a group element to itsefl is multiplication of a module element by a ring element. That is not necessarily the same as multiplication of two group elements.

Consider the additive group $\mathbb{Z}_3$ = {0,1,2} such that 1+2=2+1=0, 1+1=2, 2+2=1, 0+0=0, 1+0=0+1=1, 2+0=0+2=2.
Note that we could swap the roles of 1 and 2 in the above and it would remain an abelian group.

So can put either of the following multiplicative structures on it:

For both structures 0.x=0

Structure A: 1.x=x.1=x, 2.2=1. This is the multiplicative structure of $\mathbb{Z}_3$ qua field.
Structure B: 2.x = x.2=x, 1.1=2. This is the struture we get by swapping the roles of 1 and 2.

The two fields are isomorphic via the map 1<-->2, 0-<-->0.

If this reasoning is correct then it means that there can be more than one multiplicative structure on at least one such group, and hence the isomorphism is not always trivial.

6. Sep 6, 2015

### Dragonfall

Yes, but I need integer multiplication and field multiplication to satisfy the equation above.

7. Sep 6, 2015

### andrewkirk

I didn't say we can't have them. They come for free with every abelian group. It is standard in algebra to treat an abelian group as a module over the integers. The equation you wrote above is valid and can be proven simply by using (1) the commutativity of abelian group operations (which is denoted by the '+' symbol in this case) and (2) the distributive law for the ring of integers.

What I was saying is that multiplication of a group element by an integer (which is multiplication of a module element by an element of the over-arching ring) is not the same thing as multiplication of one group element by another, and the example in post 5 demonstrates why.