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What is the difference between a field and a group in algebra?

Could someone briefly explain it?

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Have you read the definitions of the two kinds of objects? –  Mariano Suárez-Alvarez Feb 24 '11 at 2:05
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4 Answers

This is more of a rejoinder to PEV's response as it existed before the edit. One can think of "subtraction" as an operation separate from addition, though rather than consider the binary operation, one usually considers the unary operation that maps an element to its additive inverse (it has nicer properties than subtraction, which is not associative, and it automatically captures the idea that the additive inverse is unique).

In fact, from the point of view of Universal Algebra, we want a group to have three operations: a binary operation $\cdot$ (that corresponds to the group product), a unary operation ${}^{-1}$ (that maps every element to its inverse), and a zeroary operation $e$ (distinguished element). Then the "axioms" of a group can be stated as identities of the operations:

  1. $(a\cdot b)\cdot c = a\cdot(b\cdot c)$;
  2. $e\cdot a = a\cdot e = a$;
  3. $a\cdot a^{-1} = a^{-1}\cdot a = e$.

Being expressible as identities has all sorts of nice consequences. Consequently, universal algebraists will consider a group as a set together with three operations.

Fields are more problematical, though. You start the same way: with two binary operations $+$ and $\times$; a unary operation $-$; two zeroary operations $0$ and $1$, and the axioms that describe these operations. Multiplicative inverses, however, are not defined via an operation, because $0$ does not have a multiplicative inverse. Instead, they must be defined via a partial operations, so fields are not "algebras" within the sense of universal algebra, but rather they are partial algebras. We know that fields cannot be defined as algebras with operations, because if they could then the category of fields (or even of fields of a given characteristic) with field morphisms would necessarily have categorical products, and they do not.

The theory of partial algebras is far more difficult than the theory of algebras. I seem to recall Brian Davey and David Clark, in their book Natural Dualities for the Working Mathematician mentioning that one can prove a lot of very nice and powerful "theorems" about partial algebras if one forgets some of the problems (e.g., if $f\colon A\to B$ is a homomorphism of partial algebras, then $f(A)$ need not be a partial-subalgebra of $B$; though this is particularly problematical when one consider relational algebras and relational partial algebras, where we have relations in addition to operations).

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@PEV: No, in this context, groups have three operations, period. Morphisms are required to take the image of a product into a product of images; the image of the inverse into the inverse of the image, and the distinguished element to the distinguished element. Where do you get "2 operations"? –  Arturo Magidin Feb 24 '11 at 4:13
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@PEV: For instance, in the context of monoids, that have two operations (a binary operation and a zeroary operation), morphisms are required to take the identity to the identiy (these are the "monoid morphisms", as opposed to semigroup homomorphisms between monoids). –  Arturo Magidin Feb 24 '11 at 4:17
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@PEV: It doesn't matter. If you are dealing with groups as universal algebras, you have to respect all the operations. The fact that a semigroup homomorphism between groups happens to be a group homomorphism is a special situation that has to do with groups and the identities they satisfy, not with the general set-up. (As an aside: there is a way of defining "groups" with a single ternary operation, but in that case you don't have a nice functor to semigroups and back). And, no, not "similar to an isomorphism." Please read up a bit on universal algebra; e.g., Bergman's book is a good intro. –  Arturo Magidin Feb 24 '11 at 4:44
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@PEV: The problem with what you write, by the way, is that if you view groups as having only two operations, multiplication and identity, then you cannot even write the equation $f(x^{-1}x) = f(x)^{-1}f(x) =1$, because this refers to all three operations; if you don't have the "distinguished element operation", you cannot write it down. There are inverse semigroups that are not groups, after all. –  Arturo Magidin Feb 24 '11 at 4:46
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@Andrew: You can always use "nullary" instead... –  Arturo Magidin Mar 2 '11 at 21:14
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I feel this question is too broad for MSE, and could easily be answered by using a common search engine. Briefly: A field has two operations and is always commutative. See Definiton of Group and Definition of Field.

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Looking at PEV's response, I think of division an multiplication as the same operation, similarly for subtraction/addition. –  Eric Naslund Feb 24 '11 at 2:09
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Another attempt to answer this question in more basic terms:

Groups and Fields are both important subjects of contemporary algebra. They are now recognized as strongly related, and arising in many situations :

  • fields contain two different groups as substructure that work together nicely
  • so, in order to prove that something is a field, you can first try to prove it contains these two groups, then prove other further properties (such as distributivity).

But originally, they came up from different questions.

  • Groups arose from the study of substitutions (permutations) of symbols motivated by the theory of equations and the study of symmetry of geometric figures. They were then recognized as present in many aspects of mathematics.

  • Fields arose as the generalization of the properties of the real numbers in regard to addition and multiplication, the usage of coordinates, and the expressive powers of formulas using symbols and basic operations.

But now mathematicians see them as both fundamentally linked together and to the ideas of number systems, sets of solutions of equations, solvability, the general concept of space, equations of physics. One particular kind of object which is very accessible to the beginner and can be an introduction to the links between the two are called: Finite Fields and one way to construct them is from the idea of the remainder of the division of an integer by a prime: in $29 = 7*4 + 1$, $1$ is called the remainder of the quotient of $29$ by $7$.

There is a broader concept relative to fields, called a ring, and there is the more global concept of algebra or algebraic structure which covers groups, rings and fields and many more. See also Varieties.

To have a broader idea of this family, you can lookup Lie algebras, Jordan algebras, Hopf algebras, von Neuman algebras, and further study a branch of algebra called Universal Algebra.

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Groups actually arose in the study of equations. I don't know in what sense ring is a parent concept to groups, really. –  Mariano Suárez-Alvarez Apr 25 '11 at 15:57
    
My formulation -- I will try to edit it, may give a false impression. Groups actually arose first under this name both in the study of equations designed and used by Galois as "Groups of substitutions" operating on symbols and by Cauchy as well, at the same time, operating also on points of geometrical figures designed by letters. –  ogerard Apr 25 '11 at 16:09
    
@Mariano: About rings, the problem between us is perhaps the meaning of "parent" here. I see at least two reasons to consider them close parents : one is the embedded group in a ring and another one is the fundamental way to see a finite group for instance, as a $\Z$-module or a $F_q$-module. It is the basis of modern proofs of many important results of group theory. –  ogerard Apr 25 '11 at 16:22
    
You probably mean "relative" instead of "parents" ("being a parent of" is not a symmetric relation, so that saying that two concepts are "close parents" does not make a lot of sense :) ) –  Mariano Suárez-Alvarez Apr 25 '11 at 16:27
    
In any case, that use of "parent" is rather strange. Topological spaces are also on the basis of many important results of group theory, yet it would be quite strange to say that groups and topological spaces are "close parents". –  Mariano Suárez-Alvarez Apr 25 '11 at 16:29
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As Arturo points out, "a group has three operations with signature (2,1,0) (binary, unary, zeroary), while a field has two binary, one unary, two zeroary operations and one unary partial operation" (5 operations and 1 partial operation). A group has to satisfy (i) closure (ii) associativity, (iii) identity and (iv) invertibility. A field has to satisfy these as well and also have distributivity of multiplication over addition and commutativity. We can also talk about field extensions and group extensions. As Tao notes, one usually talks about field extensions as embedding a field $F$ into a larger field $K$ that has $F$ as a subspace. Group extensions, on the other hand, are defined by covering. That is, we cover a group $G$ by another group $G'$ that has $G$ has a quotient.

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If you are going to declare subtraction and division "operations" on a field separate from addition and multiplication (which I think is terrible, pedagogically), why doesn't a group have a second operation taking $a$ and $b$ to $ab^{-1}$? For that matter, why doesn't a field have a fifth operation, taking $a$ and $b$ to $a^5-b$? Those are operations, too... unless you feel like one should "rule out" operations that aren't essential to the definition. Like subtraction and division, both of which can be characterized using only the axioms for addition and multiplication. –  Zev Chonoles Feb 24 '11 at 2:17
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While you can argue that subtraction is an operation (or better, following the universal algebra model, that "additive inverse" is a unary operation), you'll have a hard time with division (or multiplicative inverse), since it is at best a partial operation, not an operation. And if additive inverse is an operation for fields, then it's an operation for groups. If you really want to be a stickler, a group has three operations with signature (2,1,0) (binary, unary, zeroary), while a field has two binary, one unary, two zeroary operations and one unary partial operation. –  Arturo Magidin Feb 24 '11 at 3:05
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"Closure" is subsumed under "operation". –  Arturo Magidin Feb 24 '11 at 3:16
    
@Zev: In fact, the operation taking $a$ and $b$ to $a^5-b$ is a "derived operation", a member of the clone of the ring. But probably not something to introduce when one is introducing groups and rings/fields, for sure... –  Arturo Magidin Feb 24 '11 at 3:32
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