Existence of finite, non-associative group-like structures

Question

Do there exist sets $G$ with a composition such that

$G$ is finite.
There is a two-sided identity element $e\in G$ such that $eg = ge = g$ for all $g\in G$.
Each $g\in G$ has a unique two-sided inverse $g^{-1}$ with $gg^{-1} = g^{-1} g = e$.
For all $g$ and $h$ in $G$ there exists a $k$ in $G$ such that $gk=h$ (in other words: the composition table resembles a game of Su Doku)
Composition in $G$ is not associative.

I'm asking because I haven't been able to find any. If I fill out a composition table with $n$ elements satisfying 1.-3., I always end up with something associative. Even if it is possible to find these finite "non-associative" groups, it would seem that there are "many" associative ones. Hence it might be possible to formulate a criterion for when composition is associative.

Perhaps a condition like "For any $x$ and $z$ there exists a $y$ such that $xy=z$" (and possibly a similar one for right-multiplication) might be enough, though I doubt it. — Milo Brandt, Oct 05 '14 at 14:27
Your condition 4 is closure, I don't think it's what you wanted. — Najib Idrissi, Oct 05 '14 at 15:14
Yes, I mistook $k$ for $h$. Now I think the question is correct. — Gaussler, Oct 05 '14 at 18:13
@Gaussler: Do you have some application in mind? (Do you understand the importance of associativity in theory?) — C-star-W-star, Jan 26 '15 at 21:03

score 2 · Answer 1 · answered Jan 26 '15 at 20:55

Here is an example of a set X = {O,P,Q,R,S,T} with a binary operation * which is commutative and satisfies A * (A * B) = B for all A,B in X.

* | O P Q R S T
--+------------
O | O T R Q S P
P | T P Q S R O
Q | R Q P O T S
R | Q S O R P T
S | S R T P O Q
T | P O S T Q R

Now define the operation + by A + B = O * (A * B). It is easy to show that + is commutative, that A + O = A for all A in X and that each element has a unique inverse. However for this example + is not associative since (P + Q) + R = Q, while P + (Q + R) = P.

+ | O P Q R S T
--+------------
O | O P Q R S T
P | P T R S Q O
Q | Q R T O P S
R | R S O Q T P
S | S Q P T O R
T | T O S P R Q

More generally it is easy to show that if we have a set X with binary operations * and + as defined above, then (X,+) is a group if and only if A * (B + C) = C * (A + B) for all A,B,C in X.

score 1 · Answer 2 · answered Oct 05 '14 at 14:20

1

Try this:

$G$ has commutative operation $\star$, on $\{e,a,b,c\}$, defined via:
$e\star x=x$
$a\star a=b\star b=c\star c=e$
$a\star b=a\star c=b\star c=c$

Then we have $$(a\star b)\star b=c\star b=c\neq a=a\star e=a\star(b\star b)$$

answered Oct 05 '14 at 14:20

vadim123

82,796

Okay, I guess you're right. This leads me to reformulate the question, since I see that it did not quite catch all my points. – Gaussler Oct 05 '14 at 15:08
Perhaps you should close this question, think carefully about what you really want to ask, and then later ask that question separately. – vadim123 Oct 05 '14 at 15:34

Existence of finite, non-associative group-like structures

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