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Scott defines group in his book "Group Theory" as follows:

Definition: A group is an ordered pair $(G,\circ)$ such that $G$ is a set, $\circ$ is an associative binary operation on $G$, and exists $e\in G$ such that

(i) if $a\in G$, then $a\circ e=a$,

(ii) if $a\in G$, then exists $a^{-1}\in G$ such that $a\circ a^{-1}=e.$

In the Exercise 1.2.16 he asks to prove that 1) if $(G,\circ)$ and $(H,\circ)$ are groups, then $G=H$. This means that the set $G$ of a group is determined by the operation $\circ$ of the group. This fact permits one to define a group as an operation $\circ$ with certain properties. Make this definition.

I am confused! I don't know what I am supposed to do here. $\circ : G\times G\rightarrow G$ so his domain is $G\times G$. So $G$ must be equal to $H$. Did I get it wrong? How to solve the last part?

Thanks for your help.

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Really? $(\mathbb R,+)$ and $(\mathbb Q, +)$ are groups, and.... –  user21436 Jan 26 '12 at 20:46
    
@Kannappan: But is that the same $+$? –  Dejan Govc Jan 26 '12 at 20:47
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@Kannappan: well if you are really precise, I'd say one is defined as a mapping from $\mathbb{Q}\times\mathbb{Q}$ to $\mathbb{Q}$ and the other is a mapping from $\mathbb{R}\times\mathbb{R}$ to $\mathbb{R}$, so one $+$ is a restriction of the other. But I'd say that depends much on how you define things. –  Dejan Govc Jan 26 '12 at 20:51
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@KannappanSampath It depends on the definition of binary operator, but, in general, two functions with different "co-domains" cannot be "the same." –  Thomas Andrews Jan 26 '12 at 20:51
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@KannappanSampath: They are not the same as functions: they have different domains, so they are different as sets (of ordered pairs). –  Arturo Magidin Jan 26 '12 at 20:54

1 Answer 1

up vote 5 down vote accepted

If you don't want to go too deeply into the underlying set theory, note that for $(G,\circ)$ to be a group, then $\circ$ is a function with $\mathrm{Im}(\circ) = G$ by Axiom (i); and if $(H,\circ)$ is also a group (with the same $\circ$), then by the same argument shows that $\mathrm{Im}(\circ) = H$. Hence, $G=\mathrm{Im}(\circ)=H$.

If you do want to try to justify this set-theoretically, you just need to give an explicit construction of $\mathrm{Im}(\circ)$, which will require you to give an explicit description of "function", "order pair", etc in terms of sets.

In order to define groups in terms only of an operation, you need to find a way of stating the axioms using only the fact that $\circ$ is a binary operation, without referencing $G$ first. You can define the set in terms of $\circ$; or you can say things like "there exists $e$ such that for all [appropriate] $x$, $(x,e,x)\in\circ$" (where you'll have to fill in what "[appropriate]" means). And so on.

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Defining an algebraic structure entirely in terms of it's binary operations is sort of the whole point of universal algebra,Arturo.But admittedly-it is kind a strange request in a basic group theory course. –  Mathemagician1234 Jan 26 '12 at 21:00
    
I would appreciate a detailed answer, Prof. Arturo. I am feeling dumb now. Can you get to your usual style of writing long and detailed answer, please? It's a kind of unnecessary confusion, I feel. I'd like to see what those 'new' axioms would look like. I have a set of axioms in my mind, I can verify them if you write them out for me! –  user21436 Jan 26 '12 at 21:00
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@KannappanSampath: It really depends how deeply into set theory you want to delve. But you can, set-theoretically, verify that $\circ\subseteq (\mathrm{Im}(\circ)\times \mathrm{Im}(\circ))\times \mathrm{Im}(\circ)$; that for every $x,y\in\mathrm{Im}(\circ)$ there exists $z$ such that $(x,y,z)\in \mathrm{Im}(\circ)$; that for all $x,y,z,w,t\in\mathrm{Im}(\circ)$, if $(x,y,z),(z,w,t)\in\circ$, then there exists $u$ such that $(y,w,u),(x,u,t)\in\circ$ (associativity); for all $x\in\mathrm{Im}(\circ)$ there is $y\in\mathrm{Im}(\circ)$ with $(x,y,e)\in\circ$. –  Arturo Magidin Jan 26 '12 at 21:10
    
@Mathemagician1234: In universal algebra you usually talk about a set of operations, but you keep the underlying set in mind as well. Though the operations play the central role, the underlying set is "still around". I would lay it more at the feet of category theory, myself. Scott's book is from the early 60s, and there was a nontrivial sector of group theorists that were "into" categorical and general algebra thinking. –  Arturo Magidin Jan 26 '12 at 21:13
    
@KannappanSampath Ask yourself what a function is. In general, a function is a triple $(X,Y,F)$, where $X$ is the "domain" of the function, $Y$ is the "co-domain" of the function, and $F$ is a subset of $X\times Y$ with some additional conditions. In general, two functions, $(X,Y,F)$ and $(X',Y',F')$ are equal if and only if $X=X', Y=Y',\text{ and }F=F'$. A binary operator on $X$ is a of the form $(X\times X,X,F)$. In particular, if two binary operations, one on $X$ and one on $Y$, are equal, then $X=Y$. –  Thomas Andrews Jan 26 '12 at 21:23

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