Does a set $G$ need to be closed under a binary operation $\star$ for $(G, \star)$ to be a group?
Definition (Binary Operation): A binary operation, on a set $G$ is a function $$\star : G \times G \to G$$
Definition (Group): A group, is an ordered pair $(G, \star)$ where $G$ is set, and $\star$ is a binary operation on $G$ satisfying the following axioms
- $(a \star b) \star c = a \star (b \star c) \forall a, b, c \in G $
- $\exists e \in G$ (called an identity of $G$) such that $a \star e = e \star a = a \forall a \in G$
- For each $a \in G$, there is an $a^{-1}$ in $G$ (called an inverse of $a$) such that $a \star a^{-1} = a^{-1} \star a = e$
If for some $a, b \in G$, $\not\exists \ c \in G$ such that $a \star b = c$. Is $(G, \star)$ still a group?
I know that to verify that $(G, \star)$ is a group all we need to do is check the axioms for a group and verify that the necessary properties (of which closure is not one), hold, which leads me to believe that $(G, \star)$ is indeed a group, even though it's not closed under $\star$.
But perhaps more importantly, I can't see why one would define a group in this way (apart from it being obviously more general). I can't see the motivation for closure to not be an axiom for a group (again apart from generality).
As must of us know from the Vector Space axioms, that $V$ is a vector space only if it is closed under both addition and multiplication, which is where this question sprung about.
Even more generally is there any reason apart from generality, as to why some mathematical structures have an axiom for closure under a binary operation (e.g. addition/multiplication etc.) and others don't?
Best Answer
Closure is most definitely part of the definition of a group.
It might be that in some texts, closure is part of the definition of the binary operation, e.g., implicit when writing $\star : G \times G \to G$. (Rather than reiterating it as a separate axiom.)