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 $


$\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 $+$"?

  • 1
    $\begingroup$ 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. $\endgroup$
    – smcc
    Jul 18, 2016 at 10:35
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    $\begingroup$ Definition $A$. $\endgroup$
    – 5xum
    Jul 18, 2016 at 10:36
  • $\begingroup$ See here for what is by far the simplest and best definition of the word "group". $\endgroup$ Jul 18, 2016 at 10:45
  • $\begingroup$ @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. $\endgroup$
    – xxxxxxxxx
    Jul 18, 2016 at 10:58
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    $\begingroup$ Definition B is actually (nontrivially) equivalent to "$G\neq \varnothing$", see math.stackexchange.com/questions/105433/… $\endgroup$
    – PseudoNeo
    Jul 18, 2016 at 11:00

2 Answers 2


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.

  • 1
    $\begingroup$ 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 $\mathbb{R}$ $\endgroup$ Jul 18, 2016 at 14:52

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.


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