# Why do we need to specify the need for satisfying the closure property when we know that it will always be satisfied?

Wikipedia definition of a Group:

In mathematics, a group is a set equipped with a binary operation that combines any two elements to form a third element in such a way that four conditions called group axioms are satisfied, namely closure, associativity, identity, and invertibility.

If a set is equipped with a binary operation doesn't it automatically satisfy the closure property? A binary operation on a set $$S$$ is a mapping of the elements of the Cartesian product $$S \times S$$ to $$S$$, so any binary operation on two elements will give an element belonging to the same set.

Now we can have a set equipped with a binary operation where it is not associative example $$(\Bbb R,-)$$, so it makes sense to specify that it needs to satisfy associativity in order to be called a group.

My question is why do we need to specify the need for satisfying the closure property when we know that it will always be satisfied?

• Yes, but sometimes you need to check that a supposed operation really is an operation (i.e., is closed). – Randall Apr 12 at 17:12
• consider a subgroup – J. W. Tanner Apr 12 at 17:14

A binary operation on a set $$S$$ is indeed a function from $$S\times S$$ to $$S$$.