Finite Field Structure: Prime Order Cyclic Group

[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?

 

[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?

 

[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?

 

[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.)

 

[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.

 

[Dragonfall]

'Multiplication' by repeated addition is not the same as multiplication of group elements.

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

 

[andrewkirk]

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

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.