What is the difference between a set and a group?

Question

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 ?

Answer 1

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.

Answer 2

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. :)

Answer 3

Short answer .

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.

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• 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 $<G, \star >$, 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 $<G, \star >$ 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 $G_1 = <S_1, \star_1>$ and $G_2 = <S_2, \star_2>$ :

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.