# In the definition of a group, is stating the set together with the function on it redundant?

In the definition of a group (or other similar structures) $$(G,*)$$ as being the pair of a set $G$ together with binary operation $$*:G\times G\rightarrow G$$

isn't the first component of the pair basically just redundant information? Is this done for pedagogical reasons? Or has is any logical advantages, like as to extract the set from the group by sucking out the first component. I don't know how to formally steal away the codomain from a function, althoght I think it also just involves some ordered pairs.

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To be absolutely precise a group really is an order pair $(G, \cdot)$. (If you work in model theory, you may even want $(G,\cdot, e)$ to indicate the distinguished constant.)

The point is that you want to precisely define what is a group. But you notice that saying a group is just a set $G$, or the group is just a function $\cdot : G \times G \rightarrow G$ is not correct. Groups are not functions. In fact saying that a group is a set $G$ together with a function $\cdot : G \times G \rightarrow G$ is not even correct. (This is something called a structure in the language of groups.) Group even have to satisfy certain axioms. So a group really is a pair $(G, \cdot)$ which satisfies all the group axioms.

The above is a discussion about what a group really is. However, there is a distinction between the definition of a group and how much information you need so that most people would understand you in context. If you said "$\cdot : G \times G \rightarrow G$ is a group", I am sure most people would understand this to mean that $\cdot$ is the multiplication on a set $G$ which satisfy all the group properties.

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It's funny, I'm getting a little mad here. I wanted to write down the definition of a group and the problem arose when I realized that saying "$(x,y)$ is an ordered pair if axioms" is conceptually different than saying "$(G,+)$ is a group if axioms". Because in the first case you introduce the string "$(x,y)$", which as a whole is a new name and in the second set theory already knows about $(,)$. So if I want to have one general and notation independend way of definition I have to say "$\hat G$ is a group if axioms", where axioms contains that $\hat G$ equals an odered pair such that.. – NikolajK Aug 16 '12 at 20:11

It makes it convenient to say, for example:

"$H$ is a subgroup of $G$ if $H$ is a nonempty subset of $G$ such that $(H,\ast)$ is a group.

I think omitting $G$ completely in all cases is concealing too much information!

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But if we are to get technical, we don't use $*$ for $H$, but rather the existence of a function $f: H \times H \rightarrow H$ such that if $i: H \rightarrow G$ is the inclusion map, then the restriction of $*$ to $H$, $*\vert_H : H \times H \rightarrow G$, factors as $i \circ f$. And we require that $(H,f)$ is a group. Of course, it would be silly to write that all out. But the convenient shortcut is only convenient because we allow ourselves the obvious abuse of notation. – Michael Joyce Aug 16 '12 at 19:35
@MichaelJoyce Yes, thanks for spelling out the abuse for me :) – rschwieb Aug 16 '12 at 19:45

It is redundant: By definition, a function is the data of a domain, codomain, and rule that assigns to each element of the domain exactly one element of the codomain. As such, you already have the set of interest encoded in your function $*$.

It is not redundant: Well, see the other answers.

However, in all practical uses, a group is a set first, on which there is a map $*$ satisfying a few axioms.

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It's not redundant. $\ast$ isn't an arbitrary function from some set to some other set; it's specifically a function from $G \times G$ to $G$ for some fixed set $G$, and to specify this you need to name $G$ anyway.

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Okay, so one can't just name $G$ insde the function definition? I'm not sure why. – NikolajK Aug 16 '12 at 19:16
I mean, you can, but since you're probably going to refer separately to $G$ you might as well name it separately. – Qiaochu Yuan Aug 16 '12 at 21:19