# What is the difference between a set and a group?

I'm new here probably I wouldn't have a suitable way for asking suitable questions for this website really.

In group theory , I mixed between set and group in Algebra; however, I checked both of them definition.

My question here is:

Is set=group?

I think there are a large difference since we have set theory and group theory ? Can we say for example "set is finitely generated " like group ?

• All groups are sets, but not all sets are groups. See here – Michael Morrow Apr 23 at 17:54
• Welcome to Mathematics Stack Exchange. A group is a set with a binary operation (e.g., addition or multiplication) satisfying certain properties – J. W. Tanner Apr 23 at 17:55
• To have a notion of "finitely generated" you must have a notion of a binary operation. Sets by themselves do not necessarily have this. Groups however, do. – Michael Morrow Apr 23 at 17:57
• A set is a collection of things, nothing more. It is meant to describe the things we want to talk about, e.g. $\{\,0,1,2\,\}$. But it could as well be $\{\,tree, shrubbery, coconut\,\}$. A group is a structure on a set. The set defines the group elements. But we also have an addition (or likewise a multiplication). E.g. $\{\,0,1,2\,\}$ has the addition $1 + 2 = 0, 2+2=1$; the remainders of a division by three. It is the structure which makes a group on the set of group elements. – Marius S.L. Apr 23 at 18:22

There are different ways to define what kind of object a group is, however it is never a set alone. It is usually described as a set with a binary operation (such that certain properties hold), or - to clarify the meaning of "with" - as a tuple (or pair) $$(G,\circ)$$ of a set and a binary operation (with properties). Often one speaks of "the group $$G$$" instead of "the group $$(G,\circ)$$", but that is an abuse of language; nevertheless it is very common if it is somehow clear what the operation has to be. When we speak of the group $$\Bbb R$$, we actually mean $$(\Bbb R,+)$$ (and not with multiplication as operation because that would not make a group)

Another way would be to say that a group is simply a model of the group axioms, which is a very different level of abstraction.

We could try to view a group not as a tuple but as a single "thing" as follows:

Definition. A group is a map $$f$$ with the following properties:

1. $$\operatorname{dom}(f)=\operatorname{codom}(f)\times \operatorname{codom}(f)$$

2. $$f(f(x,y),z)=f(x,f(y,z))$$ for all $$x,y,z\in\operatorname{codom}(f)$$

3. For every $$x\in \operatorname{codom}(f)$$, there exists $$y\in\operatorname{codom}(f)$$ such that for every $$z\in\operatorname{codom}(f)$$, we have $$f(z,f(x,y))=z$$.

I have however never encountered this view and made it up on the spot for this answer. :) It would be fun to express 2. and 3. in terms of the projections of the direct product to its factors instead of with elements. :)

• So a set with an n-ary operation would not be a group. Is there a term for sets associated with n-ary operations? – Galen Apr 23 at 18:25
• @galen a set with a binary operation also isn't a group; a set with a binary operation satisfying certain properties is. There's no specific term for 'set with n-ary operation' because that doesn't really provide any additional structure; the reason we talk about groups as a thing unto themselves is because the contents of the group axioms provide a lot of additional structure that allows for a lot of statements 'unique' to groups. – Steven Stadnicki Apr 23 at 18:28
• Incidentally, I may be reading too much into this, but that 'single-thing' definition of group seems surprisingly close to the definition of a group object in category theory. :) – Steven Stadnicki Apr 23 at 18:29
• @StevenStadnicki Yes, sorry I was speaking shorthandedly about groups by not mentioning closure, associativity, identity, and invertibility of the operation on the set. – Galen Apr 23 at 18:33
• @StevenStadnicki I see that the term n-ary group is sometimes used in universal algebra. – Galen Apr 23 at 18:34

A group is a set $$G$$ with an associative binary operation $$\circ: G\times G\to G$$ that is closed with respect to $$\circ$$ such that there exists an identity element $$e\in G$$ where, for any $$g\in G$$, there exists $$g^{-1}\in G$$ with $$g\circ g^{-1}=e=g^{-1}\circ g$$.

Consider the empty set $$\varnothing$$. It does not contain a single element, so, in particular, it has no identity element.

• "closed" is redundant when $\circ\colon G\times G\to G$ to begin with – Hagen von Eitzen Apr 23 at 18:24
• Indeed, @HagenvonEitzen, as is specifying that $g^{-1}$ is a two-sided inverse that is unique to $g$ (as the notation implies); nevertheless, pedagogically, it helps to include such things. – Shaun Apr 23 at 18:28

Suppose you were asked to prove that a group $$G_1$$ is one and the same group as the group $$G_2$$, that is, to prove that $$G_1 = G_2$$. It would not be enough you to prove that they have the same underlying set G. You also would have to prove that their operations are exacly the same. So a group is a set with an operation ( satisfying definite properties) The 2 things are necessary to define ( that is, to identify ) a group.

• The definition you will most often encounter is : a group is a set $$G$$ with a binary operation $$\star$$ on set $$G$$ such that....

(i) the set is closed under the operation ( and this, in fact , is already implied by the expression " operation on G" )
(ii) the operation is associative
(iii) the operation has an identity element in G
(iv) every eleemnt in G has an inverse ( for this operation) in G.

And this definition will be put formally as : let $$G$$ be a set and let $$\star$$ be a binary operation on $$G$$ such that ... , the ordered pair $$< G, \star>$$ is a group.

• This can be confusing since a binary operation on a set $$G$$ is supposed , officialy, to be a function from $$G\times G$$ to $$G$$ , and therefore, a relation from $$G\times G$$ to $$G$$, and consequently a set , since a relation is ( by definition) a set.

So here is the difficulty : if a group is an ordered pair having a set as first element and an operation as second element, a group will be an ordered pair ... of sets...

• But I think we can escape these difficulties in two ways

(1) First, when we say that a group is the ordered pair $$$$, the operation ( i.e. $$\star$$) is considered intensionnaly , as an entity satisfying some properties defined via some concepts ( associativity, identity element, inverse). So the ordered pair $$$$ is not really an ordered pair of sets, but rather an ordered pair with an extensonial "part" ( the set) and an intensional "part" ( the rules/concepts defining the operation).

(2) Second, it is meaningful in mathematics ( and maybe elsewhere) to adopt as a principle that

the essence ( definition) of an entity boils down to its identity conditions.

Now , you can totally identify and reidentify a group $$G_1$$ via its base set and the binary operation acting on it , in such a way that, if ever you encounter a group $$G_2$$ with exactly the same base set and the same binary operation , you can say for sure that $$G_2 = G_1$$. So it makes sense to say that the identity ( essence, definition) of $$G_1$$ is simply the ordered pair $$< G, \star>$$, and nothing deeper than that.

In symbols :

Suppose that $$G_1 = $$ and $$G_2 = $$ :

if $$S_2 = S_1$$ and $$\star_2=\star_1$$, then $$G_2 = G_1$$.

Note : this shows the usefulness of the " ordered pair" definition of a group; it allows to use the identity condition for ordered pairs namely :

two ordered pairs are identical just in case they have excatly the same first element and the same second element.