Isn't closure of a binary operator implied by it's definition?

Question

Consider, for example, a group $(G,\circ)$, where $\circ:G\times G\to G$. Then $\text{Im}\circ\subseteq G\implies \forall a,b\in G:\circ(a,b)\in G$, giving us the closure property. Is this true? If yes, why in all group like structures asks for closure if it is implied by the operator? Where have I gone wrong, if anywhere?

Answer 1

This does seem strange to those of us following the practice of typing functions; if we say $\circ : G \times G \to G$ there is indeed nothing left to say.

However, there is an alternate practice that omits any notion of codomain. For those following this practice, the habit¹ is to simply specify "$\circ$ is a binary function on $G$", and the fact that $G$ contains all of the values of $\circ$ (when given elements of $G$ as arguments) is additional information that needs to be specified.

This may seem less strange to us if we consider a common situation mentioned in the comments; e.g. we may have some associative operation $\circ : X \times X \to X$ and some subset $G \subseteq X$, and we want to consider $(G, \circ)$ as a semigroup². By restriction we've specified a binary ($X$-valued) operation, and we need to additionally insist that it is $G$-valued.

_{1: I do not really understand this practice, so I may be misrepresenting}

_{2: Group structure has three components: the product, the inverse and the unit! If you have a group and omit the inverse and unit, you merely have a semigroup with additional properties. It's a strange feature of groups that this omission doesn't really change anything.}