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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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    @Thomas: More to the point, different domain/different image. If you view functions as sets of ordered pairs for which $(a,b),(a,c)\in f$ implies $b=c$, then a "codomain" is simply any set containing the image.2012-01-26

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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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    @KannappanSampath,@Thomas: From a set-theoretic perspective,your definitions of function are equivalent.In fact,the first rigorous definition of function I ever saw was in undergraduate group theory and it went as follows: "Let$(X,Y,F)$be an ordered triple where$X$and$Y$are sets and$F$is a nonempty subset of$X$x$Y$where no 2 different ordered pairs have the same first member. Then$(X,Y,F)$is called a function from$X$into Y."It still fascinates me as a complete encapsulation of the idea of function-but I get strange looks if I state it in front of category theorists.2012-01-28