# Does a set $G$ need to be closed under a binary operation $\star$ for $(G, \star)$ to be a group?

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

1. $$(a \star b) \star c = a \star (b \star c) \forall a, b, c \in G$$
2. $$\exists e \in G$$ (called an identity of $$G$$) such that $$a \star e = e \star a = a \forall a \in G$$
3. 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?

• Closure is also a group axiom, even if in disguise. Care to share the exact wording of group's definition you're working with? Commented Dec 23, 2016 at 23:19
• @GitGud, I've added the definitions I'm working with (out of Dummit and Foote) to the OP Commented Dec 23, 2016 at 23:35
• Given that definition, some of the answers below already effectively answer your question. To reiterate, in order to evaluate whether an order pair $(G, \star)$ is a group, it is first necessary that $\star$ is a binary operation on $G$ (hence, by definition, closed). Commented Dec 23, 2016 at 23:38

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.)

• Never have I see a definition of binary operation which isn't closed. Commented Dec 23, 2016 at 23:31
• @GitGud scalar multiplication on a vectorspace sometimes also goes under the umbrella binary operation.
– quid
Commented Dec 23, 2016 at 23:46

A binary operation $\ast$ on $G$ is a mapping $\ast:G\times G\to G$ and so by definition $G$ is closed under $\ast$. Perhaps you are thinking of verifying the axioms for subgroups, in which case you do have to verify closure. That is, to verify that $H$ is a subgroup, you have to check that the restriction of $\ast$ to $H\times H$ is a mapping $H\times H\to H$.

• Same as I said in another answer, the OP didn't say it was a binary operation on $G$. Commented Dec 23, 2016 at 23:27

part of the definition of what a binary operation is ensures closure.

Recall that the first requirement is that the group has a binary opertation, in other words a function $G\times G \rightarrow G$,

• What if the OP considers something like the usual product in $\mathbb Z$ and $G=\mathbb P$? Nothing in the question excludes this possibility. Commented Dec 23, 2016 at 23:22