What is the correct definition of a group?

Question

What is the correct definition of a group? More precisely the predicate "being a group"? According to Wikipedia

A group is a set, G, together with an operation • (called the group law of G) that...

How should one interpret this?

$\textbf{Definition A)}\\ \quad \quad G \text{ is a set},\\ \quad \quad +:G\times G\to G \\ \langle G,+\rangle \text{ is a group} :\iff\\ \quad \quad +\text{ is asscociative},\\ \quad \quad \exists 0\in G : \forall x\in G:x+0=0+x=x \text{ and } \exists y:x+y=y+x=0 $

or

$\textbf{Definition B)}\\ \quad \quad G \text{ is a set}\\ G \text{ is a group} :\iff\\ \quad \quad \exists +:G\times G\to G:\\ \quad \quad \quad +\text{ is asscociative},\\ \quad \quad \quad \exists 0\in G : \\ \quad \quad \quad \quad\forall x\in G:x+0=0+x=x \text{ and } \exists y:x+y=y+x=0 $

And is there a separate notion of "$G$ being a group with operation $+$"?

The operation can be denoted by whatever you want "$+$", "$\cdot$", or whatever. Your definition B is not correct (Definition A is). The group is defined by a set together with an operation. Using definition $B$ for any set $G$ you could define two different operations satisfying the conditions and by the definition they would be the same group. Reading that Wikipedia page should clear up your confusion. — smcc, Jul 18 '16 at 10:35
See here for what is by far the simplest and best definition of the word "group". — goblin GONE, Jul 18 '16 at 10:45
@goblin I don't think that is a good definition. I think my main problem is, the identity element does not need to be denoted by $1$, nor does the group operation need to be juxtaposition. This kind of takes away the big point of a group being an abstract structure; what we call the operation, or what we use to denote the identity element, is not important. — xxxxxxxxx, Jul 18 '16 at 10:58
Definition B is actually (nontrivially) equivalent to "$G\neq \varnothing$", see http://math.stackexchange.com/questions/105433/does-every-set-have-a-group-structure — PseudoNeo, Jul 18 '16 at 11:00
@MorganRodgers, but those are just largely irrelevant notational details. The nice thing about the linked definition is that it (0) avoids all those pointless existential quantifiers that trip up newcomers and make straightforward trivialities of group theory appear tricky and difficult, (1) fits with how varieties of algebraic structures are defined in universal algebra, (2) yields the right definition of "group object", and (3) generalizes correctly to higher-category theory. Ergo, no matter where someone is in their education, I think its always the better definition. — goblin GONE, Jul 18 '16 at 11:03
@goblin The fact that they are irrelevant notational details just reinforces the fact that they do not belong as part of the definition. For example, under this definition, it is confusing to talk about the group of integers, because you do have an element denoted by 1, but it does not satisfy the property you require. — xxxxxxxxx, Jul 18 '16 at 11:06
@MorganRodgers, there's the ghost of a good argument there, but note that, if you buy that argument, then be very careful not to write ring addition additively or ring multiplication by juxtaposition. You have to change all your notation! — goblin GONE, Jul 18 '16 at 11:07
@goblin I don't understand the problem writing ring addition additively, or ring multiplication by juxtaposition. The standard definition of a group allows you to define the operation as anything you want. Your definition seems to require that it only be written as juxtaposition. — xxxxxxxxx, Jul 18 '16 at 11:09
@MorganRodgers, oh. Now I understand what you're saying. All I can say is: maybe. Maybe you're right. Its a bit like saying "use the notation $(3,4)$ not $(x:= 3, y:= 4)$ for the relevant point in the plane." I need to think about this more before I decide whether or not I accept it. — goblin GONE, Jul 18 '16 at 11:13
@goblin I understand, thanks for taking my comments under consideration. I understand your reasoning now about the inverse map, and I like it being included, the more I think about it. — xxxxxxxxx, Jul 18 '16 at 11:21
Arguably, the right definition of group is a quadruple $(G,0,-,+)$ satisfying the right equational laws, not a pair $(G,+)$ satisfying some existential predicates. — , Jul 18 '16 at 14:33
Note that in programming, a class is kind of like a type of objects with the defined properties and methods, which are 'part of' the class. Static properties/methods correspond to constants/predicates/functions associated with the type, and one can interpret "group" as an interface that requires the class to have the static property zero and methods mult,inv with the desired behaviour. Then groups correspond to the classes that implement that interface. Elements of the group are actually instances of the class. Also, one can overload functions. — user21820, Jul 21 '16 at 12:52

littleO · Accepted Answer · 2016-07-18T10:53:58.650

11

Definition A is the correct interpretation.

A group is a pair $(G,+) $ where $G $ is a set and $+$ is a function from $G\times G $ to $G $ satisfying certain properties.

Perhaps confusingly, the group is also called $G $ (often). So two different entities -- the group, and the underlying set -- may be referred to by the same name. For example, if someone says "$g \in G $", then here $G $ is referring to the underlying set. It would be too laborious to use different names for the group and for the underlying set.

edited Jul 18 '16 at 10:53

answered Jul 18 '16 at 10:39

littleO

51,938

1

For another example of conflating the two things: the real numbers are a group (under addition) and the real numbers excluding 0 are a group (under multiplication), but the real numbers are not a group under multiplication. So the statement "the real numbers are a group" requires either that the reader already can assume you're talking about addition or some other group operator obvious from context, or else requires that the writer finish the sentence. But given that context, we confusingly do speak as if there's a predicate "being a group" which is true of $\mathbb{R}$ – Steve Jessop Jul 18 '16 at 14:52

score 3 · Answer 2 · answered Jul 18 '16 at 10:49

You appear to be asking whether on one hand a set is a group, if an operation with the correct properties exists, or on the other hand whether the group comprises both the set and the operation.

The correct definition is that the group is set together with the operation.

What is the correct definition of a group?

2 Answers2