The closure axiom is equivalent to defining the binary operation as o:GxG->G

It depends on how they define binary operation. Most define binary operation with closure included. See Fraleigh's book for example.

Usually o is a binary operation on G we mean if a ,b are in G then we mean aob is an element of G for all elements a , b in G i.e o is closed on G. So usually o is a binary operation on G and o is closed on G means the same.

Some author define binary operation as o : Cartesian product of S with S --> K , Where K need not be same as S. i.e. binary operation o need not be closed. for example define the binary operation * : Cartesian product of S with S --> K , where S is set of all negative integers and K be the set of all positive integers is not closed . That is why they include the closure property in a group. i.e. for closure property K must be S.

The Closure Axiom is important in the definition of a group. Depending on how the binary operation is define we have to proof it or not (in the case when it is given that 'o' is a binary operation on the set G).

If ind the case said that 'o' is binary. That is no need to include clossure axiom, because clossure axiom used to proof that 'o' is binnary

But if 'o' is new operation that unknown binary or not that need to use clossure axiom to said that 'o' is binary

Yeah, No need to include the closure axiom, if the operation is binary

No need since closure is a property of the binary operation already by definition.

And if a closure property of a binary operation is to be proved then a contradiction is reached by supposing it is not and deduce a non closed but closed operation.

The definition of a group is a set G together with a function "o" from GxG to G satisfying the properties of associativity, existence of a neutral element, existence of inverses. Saying that o maps GxG to G is called "closure".

If you define o: G\times G \to G closure is build into the definition already. However, if you have a subset K \subset G which then inherits an operation o' K \times K \to G

(which is automatically associative) you have to check that the image of o' on K lands in K and so defines a binary operation o'' K\times K \to K. Likewise you have to check that the restriction of the inverse to K inv': K \to G lands in K and so defines a map inv'':K \to K (which is then automatically an inverse for o'' and gives that 1 \in K). So consider insisting on closure didactic: it if you have an associative operation o(k, k') = k * k' with unit and inverse given, you have to check that it the operation is closed, i.e. lands in K rather than in a larger space which you have conveniently suppressed in the notation *.

I think there is no need to take the closure axiom again, if it has been already defined in the binary operation because the binary operation itself says that G is closed under that operation.

Indeed no need to include closure since it is already a property of a binary operation.

If G is anon empty set and o is a map from G x G to G defined by o (a,b) exactly one element in G namely a ob .Then o is called a binary operation on G.Thus o binary operation means o is closed on G.

If o is associative and there is identity element say e s.t e o a =a o e= a for all a in G and for all a in G there is b in G s.t. a o b= b o a= e.Then G is a group.

0 is a binary operation on a set G itself says that G is closed under 0. Yes, one thing more, if we define a mapping * from G \times G to G, where G is a set of equivalence classes, we must show that * is well defined to fulfill the first step to form (G, *) a group.

A map o from g x g to g defined by o(a,b) is rule that assign to each order pair (a.b) in g x g exactly one value in g namely o(a,b) is called a binary operation.in other words a binary operation is said to be closed if for any two elements a,b in g we have a o b in g.

Thus binary operation and closed mean exactly the same thing

Let G, T be nonempty sets and let α: G→ T be a bijective mapping and let o:G×G→T be a mapping. Call (G,T,α,o) an almost group if it satisfies the following conditions:

1.There exists an element e∈G such that 0(e,g)=α(g) for all g∈G.

2. For every g∈G There exists an element g⁻¹∈G such that o( g⁻¹,g)=α(e).

3. For all g₁,g₂,g₃∈G, o(g₁,α⁻¹o(g₂,g₃))=o(α⁻¹o(g₁,g₂),g₃).

Note that if T=G and α=I_{G} the identity on G,then (G,I_{G},o) will be the usual group. How is that sound?

What you see the advantage of introducing the map alpha ? . I did not think any major example which gives interesting result with this definition of atmost group. Can you give some better example if you really think ?

Consider the group (Z₃,⊕ ).Let G=T=Z₃. Define o:G×G→T such that o(x,y)=x⊕y and let α:G→T such that α(x)=x⊕1

Then we get an almost group (G,T,α,o) with e=1 and 0⁻¹=2, 1⁻¹=1, 2⁻¹=0. However if we define α:G→T such that :

0→1, 1→0, 2→2, then the element e is not exist.

No need for this map as it will not help the reader to understand the group easily . I gave two answers previously and confirmed that binary operation and closed means the same thing.