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.

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    $\begingroup$ Presumably you mean nontrivial homomorphism? If not, sending $\phi(k) = 1_F$ is a group homomorphism from the additive group $F$ to the multiplicative group $F$. $\endgroup$ – walkar Feb 18 '15 at 2:24
  • $\begingroup$ Yes sorry I did mean non trivial $\endgroup$ – ASKASK Feb 18 '15 at 2:24

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.

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    $\begingroup$ I was planning on posting this same argument. Darn! One slight note: You don't actually need the hypothesis that $q$ is a prime power to prove this. $q$ and $q-1$ are coprime for any $q$. (I note this just because proving that the order of a finite field is a prime power requires much heavier machinery than just Lagrange's theorem) $\endgroup$ – Milo Brandt Feb 18 '15 at 2:30
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    $\begingroup$ Is there a countable field with such a homomorphism? It doesn't look like power maps do the trick on $\Bbb Q$ and $\Bbb A$, since the power of a rational number is algebraic and the power of an algebraic is transcendental (usually). $\endgroup$ – Mario Carneiro Feb 18 '15 at 2:42
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    $\begingroup$ I wondered something similar, so I created this question: math.stackexchange.com/questions/1153994/… $\endgroup$ – Qudit Feb 18 '15 at 2:54
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    $\begingroup$ @Mario To give a simple example of a countable field with a map from additive to multiplicative group, consider the set $S$ to be the smallest subfield of $\mathbb R$ closed under the map $x\mapsto e^x$. This is countable and $x\mapsto e^x$ is a homomorphism. $\endgroup$ – Milo Brandt Feb 18 '15 at 2:56
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    $\begingroup$ @AlexH It is finitely generated, so in correspondence with finite strings over a finite alphabet - every element can be written as a finite tree of applications of the unary operations of taking negatives, taking reciprocals, and exponentiation, the binary operations of addition and multiplication, and the constants $0$ and $1$. $\endgroup$ – Milo Brandt Feb 18 '15 at 4:33

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.)

  • $\begingroup$ @MorganRodgers: In the first example, the finite field A is the additive group of integers modulo 4 -- or in other words, A is the set of four integers {0, 1, 2, 3}. In the second more general example, A is the additive group of of integers modulo N-1, where N is a Fermat prime. (The first example is a special case of the second example). How could I edit this answer to make that more clear? $\endgroup$ – David Cary Jul 7 '15 at 23:26
  • $\begingroup$ That is not a finite field; the additive structure of $\mathbb{F}_{4} \simeq \mathbb{Z}_{2} \times \mathbb{Z}_{2}$. If this is representing the multiplicative structure of $\mathbb{F}_{5}$ i think there needs to be some clarification (and in this case I think you are mapping $\mathbb{F}_{5}^{*}$ to itself, using additive notation in $A$ and multiplicative in $M$). $\endgroup$ – Morgan Rodgers Jul 7 '15 at 23:55
  • $\begingroup$ @MorganRodgers: You are right that the additive group of integers modulo 4 has a completely different addition operator than GF(4), the finite field of four elements {0, 1, a, 1+a}. I don't know what I was thinking. You are probably right that I am "simply" mapping GF(5) to itself using additive notation in A and multiplicative notation in M -- but isn't that almost exactly what the original question asked for? $\endgroup$ – David Cary Jul 8 '15 at 13:38
  • $\begingroup$ No, he wanted a homomorphism from $(\mathbb{F}_{q}, +)$ to $(\mathbb{F}_{q}^{*}, \cdot)$ (which is impossible). Your map represents elements of $\mathbb{F}_{5}^{*}$ in two ways; one as the integers $\bmod{5}$ (in $M$; notice that $0$ is not in the image of your map), and once as the exponents of the generator (so essentialy, they are being represented in $A$ as ${0,1,2,3}$ precisely under the map you define). This map you define will actually work (trivially) for any prime (or prime power), where $2$ is replaced by an arbitrary primitive element. $\endgroup$ – Morgan Rodgers Jul 8 '15 at 17:44

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