I just couldn't let go of this. A Group operator is typically defined to be a single operator like a '+' or as a composition operator like in a Permutation group. However, we fail to precisely define an operator - more generally! For e.g. Let the group $G \subset \mathbb{Z}$ be a subset with the following properties.

  • Operator : The '+' operator could be defined to be $f(x,y)=x+y, f:\mathbb{Z^{2}} \to \mathbb{Z}$ being the group operator.
  • Closure: $\forall (x,y) \in G, f(x,y) = z \vert z \in G$
  • Identity: $ \exists i \in G \vert f(x,i) = i $ $\forall x \in G$
  • Inverse: $ \exists (y,i) \in G \vert f(x,y) = i$ $\forall x \in G$
  • Commutative: $\forall (x,y) \in G, f(x,y) = f(y,x)$
  • Associativity: $\forall (x,y) \in G, f(z, f(x,y)) = f(f(z,x),y)$

I think then this captures the idea the generalized version of the operators could be some function $f$ instead of some specific operation that is very narrow or more specific to the problem in hand. This could be extended to include infinite and finite sets and discrete or continuous. $f: \mathbb{R}^{n}\to \mathbb{R},$$f: \mathbb{C}^{n}\to \mathbb{C}$ which is about how general we can get.This can be then generalized more to include the search for such operator functions and the corresponding $G$ that satisfies the aforesaid conditions. The "Well known" discrete groups such as $S_{n}$ is merely a specific Group with some "fixed" operator and identity. I guess we could cover it to include Lie groups as well. Instead of searching for Groups that contain specific fixed operators we should instead focus on fixed Groups/subsets with search for these operator functions that generate that Groups/subsets and may be relations between such Groups/subsets.

Here is an example, consider a set of points in the $(+,+)$ and $(-,-)$ quadrants of a circle at center (0,0). Call it P and Q. Now under the function $f(x,y) =x +y$. Such that $x \in P$ and $y \in Q$. We can assemble a set of points from each of these quadrants such that we can form a ‘Structure’. The inverse of course exists with $(0,0)$ being the identity. Now, my question is why restrict it to one such function $x+y$. There must be other functions, in general, that apply to these points that also generate a similar structure. All I’m trying to say is shouldn’t the concept of group be studied under this general concept instead of fixing the ‘+’ or other operations

Just wondering what you guys think and would appreciate more info!

  • $\begingroup$ Note: $\Bbb Z\ast\Bbb Z$ means the free product of $\Bbb Z$ with itself, not the direct product $\Bbb Z\times\Bbb Z$ of $\Bbb Z$ with itself. $\endgroup$ – Shaun Jan 29 at 0:26
  • $\begingroup$ Somehow I think this is related. $\endgroup$ – Shaun Jan 29 at 1:12
  • $\begingroup$ I'm no that's sure what you've written is somehow different from the definition of a group (excluding commutativity of course) $\endgroup$ – leibnewtz Jan 29 at 22:06
  • $\begingroup$ By allowing the operator $f(x,y)$ to be defined as a function we can introduce for e.g $f(x,y) = e^{x+y}$ is my point. Which we wouldn't ordinarily consider to be an operator $\endgroup$ – Gopal Anantharaman Jan 29 at 22:11

I'm not sure I understand your question but here's a collection of thoughts on the opening post.

The closure axiom in group theory is often omitted by some authors as it is implied by, in a model of the theory (i.e., a group $(G, \circ)$), the binary operation being just that: a binary operation defined by

\begin{align} \circ: G\times G&\to G \\ (g_1, g_2) &\mapsto g_1\circ g_2, \end{align}

so that $g_1\circ g_2\in G$.

Nonetheless, yes, one has to check closure from time to time to see if a candidate group is indeed a group, but such cases are usually subgroups of some group or even a semigroup. I think there are some esoteric exercises or lemmas - even theorems - that start with the assumption of some closure properties that end up showing the resulting structure is a group.

One such exercise is as follows.

Consider a semigroup $S$. Show that $S$ is a group if and only if $aS=S=Sa$ for every $a\in S$.$^\dagger$

Also Exercise 39 of the supplementary exercises for chapters 1-4 of Gallian's "Contemporary Abstract Algebra" is a 1995 Putnam question. You might be interested in it. It explores closedness without explicit use of group theory.

$\dagger$ Here $aS:=\{as\mid s\in S\}$ and $Sa$ is defined dually. Reference for exercise: Howie's "Fundamentals of Semigroup Theory", somewhere near the beginning.


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