# Why is closure omitted in some group definitions?

In some texts, there are three group axioms and in some there are four. The difference is that one of the axioms, the closure ($a,b\in G$ then $a*b \in G$) is omitted. Why is this so?

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If the definition of group says that $*$ is "a binary operation", then the word "operation" already includes closure, since "binary operation on $A$" means "a function $A\times A\to A$". So the values necessarily are taken in $A$. – Arturo Magidin Sep 11 '11 at 21:26

When three axioms are given (associativity, existence of an identity, existence of inverses), these are axioms that apply to a set $G$ with a binary operation. A binary operation is by definition a function from $G\times G$ to $G$, meaning that for each ordered pair $(a,b)$ of elements of $G$, there is a unique element (denoted $a*b$ or $ab$ or some other way) of $G$ determined by the operation. Thus, the "closure axiom" is incorporated into the definition of a binary operation on a set.

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Without knowing the full context, I will guess that such texts defined $*$ to be a 'binary operation,' or some other definition that somehow encompasses closure within it but is slightly hidden. I have read a text that also assumed associativity of the binary operation, so that all that was needed was the existence of the identity and the inverse. But, defining a group with a binary operation or a group with an arbitrary operation that is closed and associative is more or less the same thing.

At the end of the day, the group operation is closed and associative, regardless of how explicit or implicit the book makes it.

I know I typed this up without knowing the exact context, but is that right? Did that book use some pre-defined or specified operator?

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Yes. Yours and Arturo's assessment is correct. The existence of the binary operation was made explicit in the definitions which did not require closure. I did not notice this as I thought that of a binary operation as $*:A\times A \rightarrow \mbox{Anything}$, or didn't think of it at all. – kuch nahi Sep 11 '11 at 21:32
Taking "binary operation" to imply associativity is just sloppy, I think -- definitely nonstandard. I wonder how that author would define an abstract Lie algebra. – Henning Makholm Sep 11 '11 at 21:36
@Henning: I completely agree. But that was long ago, and I can't remember who it was anymore. – mixedmath Sep 11 '11 at 22:01
@kuch nahi: Here's a thought. If the operation were a function from $G\times G$ to some set $B$ (with $B$ not necessarily equal to $G$), then what would associativity mean? Suppose $a$ and $b$ are in $G$ but $a*b$ is not; then if $c$ is in $G$, what is the meaning of $(a*b)*c$ when trying to state associativity? – Jonas Meyer Sep 13 '11 at 5:15

Stating closure explicitly allows the definition of a subgroup to simply be a subset that is a group when you restrict the group operation to it.

The point here is that when you restrict the operation of $G$ to a subset $H$, you get a map $H \times H \to G$, and so closure for $H$ means this map is actually $H \times H \to H$.

Assuming closure as implicit in the definition of binary operation is fine, but it does make the definition of subgroup slightly more wordy.

Of course, this is a matter of taste and different writers choose different definitions according to their taste.

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What is the first "it" referring to? I'm not sure what you mean, because this could be true with either approach. If binary operations are assumed, then part of saying that the subset with restricted operation is a group is saying that the restricted operation is an operation on the subset, which is equivalent to "closure". – Jonas Meyer Sep 11 '11 at 23:20
@Jonas, re-reading the question and my answer, it seems that I haven't really answered the question... But I wanted to make that point about subgroups... – lhf Sep 11 '11 at 23:40
lhf: Thanks for clarifying, and you make a good point (+1). – Jonas Meyer Sep 12 '11 at 2:23