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I've seen the definition of a group in many different books given as follows:

A group is a nonempty set $G$ and a binary operation $*$, denoted by $(G,*)$ that satisfies the following properties:

  1. For every $a$, $b$, and $c$ in $G$, $(a*b)*c=a*(b*c)$;
  2. There exists $e$ in $G$ such that for every $a$ in $G$, $a*e=a$;
  3. For every $a$ in $G$, there exists $b$ in $G$ such that $a*b=e$.

I would just like to clean this up and put it in predicate logic properly. I have not seen this done in any book. When I try to do this myself, I feel like I'm opening Pandora's box! Here is how I have reasoned about this important definition:

Let $G$ be any nonempty set and form the set containing $G$ as the only element, $\{G\}$. Now let $*$ be set of all functions from $G\times G$ to $G$, so $*=\{\text{all functions } G\times G\rightarrow G\}$. Next, we take the Cartesian product of $\{G\} \times * = \{ (G,*_1), (G,*_2), (G,*_3),\dots\}$. So, basically we have set of ordered pairs where each ordered pair consists of a set and a function. It seems that we are interested in whether or not each ordered pair satisfies the above properties. Actually, it seems that for a particular ordered pair $(G, *_n)$ to be a group, it would only be dependent upon how the function is defined. Depending on the cardinality of $G$, there are many functions from $G\times G\rightarrow G$, and we know that not all of these would be groups, nevertheless, it seems that $(G,*)$ is group if $*$ satisfies the above properties.

So, I've tried to formulate the law of associativity using only ordered pairs. Here is what I have come up with:

For every $a$, $b$, $c$, $x$, $y$, $z$ in $G$ ($*$ is associative iff [$((a,b),x)$ in $*$ and $((x,c),y)$ in $*$ iff $((b,c),z)$ in $*$ and $((a,z),y)$ in $*$])

I know this looks ugly, but since $*$ is really just a set of ordered pairs where the first coordinate is also an ordered pair, I tried to translate For every $a$, $b$, $c$ in $G$, $(a*b)*c=a*(b*c)$ strictly into a statement with ordered pairs.

Anyway, I'm not sure about this formulation. For example, when we want to test if a relation $R$, on $S\times S$ is symmetric, we say for every $a$, $b$ in $S$, $R$ is symmetric iff $(a,b)$ in $R$ implies $(b,a)$ in R. The difference is technically $*$ is a relation on $(G\times G)\times G$ and to test for associativity I said for every $a$, $b$, $c$, $x$, $y$, $z$ in $G$. Can I do this since for symmetry, $R$ was a relation on $S\times S$ where the two sets are equal. On $(G\times G)\times G$ the two sets are not equal, so I don't know if this is correct.

Besides the issue with associativity, I'm not sure how to formulate the last two rules of groups in predicate logic and also I think we should include something about closure, perhaps.

I just want to have one long statement in predicate logic that correctly gives the definition of a group incorporating everything in the English version given above.

Any help would be appreciated. Thanks.

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It's curious that you insist on replacing the notion of function with a representation as a set of ordered pairs, but you don't replace the notion of ordered pair with a representation as a set. – Hurkyl Sep 5 '12 at 10:47
This sounds like you're trying to create a formal description in the language of set theory that describes groups as a set-theoretical concept. Is that because you explicitly want it to be about set theory (perhaps your real goal is to understand the reduction of everything to set theory better?), or because you just aren't aware of the "native" first-order-logical approach presented in William's answer. – Henning Makholm Sep 5 '12 at 10:50
Hurkyl, yes I will replace the ordered pairs I have listed above with their set representations once I have formulated the entire definition in terms of set theory. – Josh McKinney Sep 6 '12 at 20:35
Henning, that's exactly what my goal is, but I'm not sure how to put everything together. I can break down certain pieces, but I have difficulty in putting them together. – Josh McKinney Sep 6 '12 at 20:37
@Josh: I was really hoping you'd go the other way and realize that abstraction is a Good Idea, rather than go through even greater lengths to eliminate it. :( – Hurkyl Sep 10 '12 at 10:01

The first order theory of group is described as follows:

The language of group theory is $\mathfrak{L} = \{\cdot, e\}$, where $\cdot$ is a binary function symbol and $e$ is a constant symbol.

The first-order theory $T$ of group theory consist of the following axioms :

1) $(\forall x)(e \cdot x = x \cdot e = x)$

2) $(\forall x)(\forall y)(\forall z)(x \cdot (y \cdot z) = (x \cdot y) \cdot z)$

3) $(\forall x)(\exists y)(x \cdot y = e)$

A structure in the language $\mathfrak{L}$ is a tuple $(G, \cdot^G, e^G)$ where $G$ is a set, $\cdot^G$ is a function $G \times G \rightarrow G$ and $e^G \in G$. A structure in the language $\mathfrak{L}$ is a model of $T$, the Theory of Group Theory, if and only if $G$ satisfies all the axioms of $T$. A $\mathfrak{L}$-structure $(G, \cdot^G, e^G)$ that is a model of $T$ is called a group.

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Thank you, but I was looking for a set-theoretic definition using the language of set theory. I did not want to create a new language set to define a group. Anyone know of any resources that might have what I'm looking for? – Josh McKinney Sep 6 '12 at 8:53

I think that associativity can be formulated simply using "set-theoretic" notions thusly:

$Let\space f \in \ast \space and \space G \space be \space a \space set.$

$(G, f)\space is \space associative \space\space iff \space\space \forall a,b,c \in G; f(G \times G) - \{f(a,(b,c))\} = f(G \times G) - \{f((a,b),c)\} % $

$Where \space f(x,y) \space is \space equivalent \space to \space an \space ordered \space pair \space ((x,y),z) \in f.$

Which is equivalent to:

$Let \space f \in \ast \space and \space G \space be \space a \space set.$

$f\space is \space associative \space\space iff \space\space \forall a,b,c \in G; f(a,(b,c)) = f((a,b),c) % $


$\forall a,b,c \in G; \space if \space \exists d,e \in G\space s.t. \space ((a,(b,c)),d) \in f \space and \space ((a,(b,c)),e) \in f \space then \space d = e. $

Your formulation strikes me as wrong because you specify that $\ast$ is a set of functions, i.e. its elements are sets of ordered pairs. Then you say $((a,b),x)\space in \ast$. But $((a,b),x)$ isn't in $\ast$, since that would imply that $\ast$ is a set of ordered pairs. I can't see how a set of ordered pairs (the elements of $\ast$) can be morphed into an ordered pair.

The other axioms can be formulated just as easily.

Forgive me if I'm saying something totally naive, or if I have miscontrued somehow what you want.

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Yes, I did make a mistake. I meant to write, ((a,b),x) in *_n where *_n is a element of *. It seems like that last thing you wrote above is just the definition of a function from GxG-->G. – Josh McKinney Sep 12 '12 at 5:42

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