2
$\begingroup$

I'm curious if there are any cayley tables on a finite amount of elements that satisfy the axioms of a) closure, b) identity, and c) inverse, but that for at least one triple of elements do not satisfy the associative property, and so thus the set is not a group.

I'm also wondering if there are any in which every element has a unique inverse but a group is still not formed because associativity is not held.

$\endgroup$
3
$\begingroup$

For the table $$\matrix{e&a&b\cr a&e&e\cr b&b&e\cr}$$ we have $$(ab)a=ea=a\quad\hbox{but}\quad a(ba)=ab=e\ .$$

Another example: $$\matrix{e&a&b\cr a&e&a\cr b&b&e\cr}\ ,\qquad (ab)a=e\ ,\qquad a(ba)=a\ .$$

$\endgroup$
  • $\begingroup$ This is a very great example, I just forgot to mention one thing in my question, I'm also looking for an example in which every element has a unique inverse $\endgroup$ – ASKASK Feb 10 '15 at 3:03
  • $\begingroup$ In this example every element is its own unique inverse. In particular, $a$ and $b$ are not inverses since $ba\ne e$. $\endgroup$ – David Feb 10 '15 at 4:00
  • $\begingroup$ Oh I did not notice that, thanks $\endgroup$ – ASKASK Feb 10 '15 at 4:03
  • $\begingroup$ Alternatively, see the example I just added. $\endgroup$ – David Feb 10 '15 at 4:05
  • $\begingroup$ math.stackexchange.com/questions/619891/… See here. It's a pretty interesting problem. $\endgroup$ – Cameron Williams Feb 10 '15 at 4:39
1
$\begingroup$

Consider the following Cayley Table of order $5$:

$$\begin{array}{c|ccccc} \ast & 1 & 2 & 3 & 4 & 5\\ \hline 1 & 1 & 2 & 3 & 4 & 5\\ 2 & 2 & 1 & 3 & 3 & 5\\ 3 & 3 & 3 & 1 & 4 & 5\\ 4 & 4 & 3 & 4 & 1 & 5\\ 5 & 5 & 5 & 5 & 5 & 1 \end{array}$$

We have that $\{1,2,3,4,5\}$ is closed under $\ast$, $1$ is the identity element and every element is the unique left/right inverse of itself (see the diagonal of the table).

Moreover, this operation $\ast$ is commutative too, but not associative:

$$(2\ast 3)\ast 4 = 3\ast 4 = 4 \ne 3 = 2\ast 4 = (2\ast(3\ast 4)).$$

What's interesting is that one such structure, that

satisfies all group axioms except associativity,

is not necessarily even a quasigroup. Indeed in my example, as in the first one of two in the answer by @David, not all the rows (columns) of the Cayley Table are bijections of the underlying set: such structures are indeed not cancellative and have no division (see definitions 5 to 9 in this paper).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.