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.
- 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
(iii) the operation has an identity element in G
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.