Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm wondering, which is the smallest quasigroup which is not a group? And how to check it?

share|improve this question
    
Let us say "which are the smallest quasigroups"! –  MattAllegro Apr 18 '14 at 23:00

3 Answers 3

The Cayley table:

$$\begin{array}{c|ccc} \ast & 0 & 1 & 2 \\ \hline 0 & 0 & 2 & 1 \\ 1 & 1 & 0 & 2 \\ 2 & 2 & 1 & 0 \end{array}$$ represents a finite quasigroup of order $3$ over the set $\mathbb{Z}_3$ of the integers mod$3$. The operation $\ast$ is $$a\ast b=(a-b)\text{mod}3.$$


Check that the operations $$a\bullet b=(a+b)\text{mod}2$$ and $$a\circ b=(a-b)\text{mod}2$$ over the set $\mathbb{Z}_2$ of the integers mod$2$ give raise to the same Cayley table: $$\begin{array}{c|cc} \bullet & 0 & 1 &\\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \end{array}$$

share|improve this answer

Some authors allow the empty set endowed with the empty binary relation to be a quasigroup, whereas a group must contain an identity element. This is neither satisfying nor illuminating, so we'll disallow this.

Recall that quasigroup structures $\ast$ on a finite set $\{a_1, \ldots a_n\}$ can be identified with $n \times n$ Latin squares filled with the symbols $a_k$, simply by taking the Latin square to be the multiplication table. For order $1$ there is only one multiplication table, and this defines the group structure on the trivial group.

For order $2$ there are only two Latin squares, namely $$ \begin{array}{c|cc} \ast & a & b \\ \hline a & a & b \\ b & b & a \\ \end{array} \qquad \text{and} \qquad \begin{array}{c|cc} \star & a & b \\ \hline a & b & a \\ b & a & b \\ \end{array} $$ Checking directly shows that the transposition $(ab)$ is an isomorphism $(\{a, b\}, \ast) \to (\{a, b\}, \star)$, so again up to isomorphism there is a unique quasigroup of order $2$, and this must be the group of order $2$.

However, for order $3$, there are $12$ Latin squares leading to five isomorphism types of quasigroups, and just one of these is a group structure (and in fact only this one has an identity element), so as MattAllegro writes in the comments, one should really be asking about the smallest quasigroups that are not groups. (The number of quasigroup isomorphism types is the content of OEIS A057991.)

The multiplication table for the group, $(\mathbb{Z}_3, +)$, is $$ \begin{array}{c|ccc} + & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 3 \\ 2 & 2 & 0 & 1 \end{array} . $$ Representatives of each of the remaining four quasigroup isomorphism types of order three (that is, precisely the smallest quasigroups that are not groups, up to isomorphism) are specified by the following multiplication tables: $$ \begin{array}{c|ccc} - & 0 & 1 & 2 \\ \hline 0 & 0 & 2 & 1 \\ 1 & 1 & 0 & 2 \\ 2 & 2 & 1 & 0 \end{array} \qquad \begin{array}{c|ccc} \ominus & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 2 & 0 \end{array} \qquad \begin{array}{c|ccc} \ast_1 & a & b & c \\ \hline a & a & c & b \\ b & c & b & a \\ c & b & a & c \end{array} \qquad \begin{array}{c|ccc} \ast_2 & a & b & c \\ \hline a & b & a & c \\ b & a & c & b \\ c & c & b & a \end{array}. $$

Here, $-$ denotes subtraction modulo $3$, and $\ominus$ denotes subtraction modulo $3$ with the arguments in the reverse order, namely, $x \ominus y := y - x$.

Note that $\ast_1$ is determined among all quasigroups on $3$ elements by idempotency (that is, by the property that $x^2 = x$ for all $x$), and $\ast_2$ is determined (up to isomorphism) by commutativity and the absence of an idempotent element.

One can, by the way, enumerate these isomorphism types and produce explicit multiplication tables in Maple with the command

Magma:-Enumerate(3, 'quasigroup', 'output' = 'list');
share|improve this answer

Looking at Wikipedia: Small Latin squares and quasigroups it is quite clear that quasigroups of order two or below are really groups, while for order three there is a quasigroup with no identity element.

(Of course if you allow the empty set with its unique binary opeation as a quasigroup, that structure is not a group.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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