I used "elements" in the title because I know that technically $(S, \cdot)$ is a set with two elements.

Consider the set $S=\{e, a_1, a_2, a_3, ... \}$ and define the operation $\ast:S \times S \to S$ by $x \ast y = e$ for every $x, y \in S$. Then $\ast$ is a binary associative operation on $S$, and it is not difficult to show that the group axioms hold for $(S, \ast)$. But in this group, every element of $S$ behave as the identity; and since the identity is unique, we must $e=a_1=a_2=a_3=\cdots$ and thus $G$ is the trivial group of order $1$, even though in the set $S$ we have $e \not = a_1 \not = a_2 \not = a_3 \not = \cdots$ .

Is my argument above correct? If so, I have a couple of follow-up questions:

$1)$ Is there some terminology which distinguishes between the elements of the set $S$ and the "elements" of the group $G$?

$2)$ Why do we seem to never worry about this in group theory books? In every book that I've seen, we seem to define a set of $n$ elements, define a binary operation on it, and assume that the group also has $n$ "elements".

EDIT: Sorry if the title confused anybody, I had "more" when it should've been "less".

  • $\begingroup$ Your argument shows that no matter how we set up the binary operation, we can always throw out all the extra elements to get what we would call normally a group, I think. $\endgroup$ – Alfred Yerger Dec 23 '17 at 18:16
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    $\begingroup$ 1: $(S,\cdot)$ is defined to be a set with $|S|$ elements, and an operation. Not necessarily with 2 elements. 2: The group axioms do not hold for $S$, unless $S = \{e\}$. 3:: We don't have every element acting as the identity (again unless $|S| = 1$), I don't understand why you have that impression. 4: What are you trying to show? $\endgroup$ – Morgan Rodgers Dec 23 '17 at 18:19
  • $\begingroup$ @MorganRodgers While trying to answer question $2$ I realized that you are right and I wanted to delete this question (before it got an answer). But just to address $1$, in my textbook (Dummit & F) a group is defined as an ordered pair and ordered pairs are sets with two elements. $\endgroup$ – Ovi Dec 23 '17 at 18:43
  • $\begingroup$ An ordered pair is not exactly a set with two elements, I would be careful thinking like this. Having an ordered pair means there are two objects in a particular order, but it is really a different from being a set. $\endgroup$ – Morgan Rodgers Dec 23 '17 at 18:57
  • $\begingroup$ @MorganRodgers Hmm I don't think of ordered pairs intuitively as sets, but I thought formally the ordered pair $(a, b)$ is defined as the set $\{\{a\}, \{a, b\} \}$ en.wikipedia.org/wiki/Ordered_pair#Kuratowski's_definition $\endgroup$ – Ovi Dec 23 '17 at 19:03

Say you have two distinct elements $x,e\in S$, where $x\neq e$. If $(S,\cdot)$ were to satisfy the group axioms then we would have $$x = x\cdot e = e.$$ which is a contradiction. Actually, you should rewrite $\cdot$ in function notation: $$\cdot(x,e) = x$$ $$\cdot(x,e) = e$$ So $\cdot$ as a function is not well defined. (Does it send $(x,e)$ to $x$ or to $e$? These are different results by assumption.) Thus it is not a binary operation on $S$.


Your binary operation $*$ does not satisfy the group axioms (assuming that $e,a_1,a_2,\dots$ are all distinct elements of $S$)! In particular, it has no identity element: there is no element $x\in S$ such that $x*y=y*x=y$ for all $y$. Indeed, it is impossible to have $x*y=y$ unless $y=e$.

More generally, the distinction between elements and "elements" you are trying to make does not exist. Your argument, if correct, would simply give a contradiction, since you have both $e=a_1$ and $e\neq a_1$, for instance.


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