Given any finite group with multiplication $m(-,-)$ and three permutations $p,q,r$ on the underlying set of the group, we can obtain a quasigroup with binary operation $g*h:= p(m(q(g),r(h)))$. What is an example of a finite quasigroup that provably does not arise in this way?
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asked Apr 1, 2014 at 15:37
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$\begingroup$ You could look at Latin squares up to permutation. OEIS entry oeis.org/A123234 suggests there are 4 such of order 4, from which two of them are not isotopic to a group. Bruck-Toyoda suggests such a loop is not medial. Gerhard "Web Searches Can Be Fun" Paseman, 2014.04.01 $\endgroup$– Gerhard PasemanCommented Apr 1, 2014 at 16:03
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1 Answer
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The construction is called "isotopy" and you ask for quasigroups that are not "isotopic to groups".
It's been proved a long time ago that if a loop (= a quasigroup with a unit element) is isotopic to a group G, it is actually isomorphic to G. Consequently, any non-associative loop is an answer to your question. The smallest example has five elements, and is unique (up to isotopy):
0 1 2 3 4
1 0 3 4 2
2 3 4 1 0
3 4 0 2 1
4 2 1 0 3
The same theorem gives a simple criterion to recognize quasigroups isotopic to a group. Given a quasigroup with an operation *, pick an element e and form a new quasigroup, by x # y = x/e * e\y. This is a loop, where e*e is the unit element. If # is associative, then, obviously, * is isotopic to a group. If # is not associative, the theorem assures no group isotope can be found.
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$\begingroup$ Thanks! Given that this is mathoverflow, could you actually give the example of this 4-element quasigroup. Then I will accept the answer. $\endgroup$ Commented Jan 21, 2015 at 21:47
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$\begingroup$ Did I say four? Correct is five, sorry! Table added. $\endgroup$ Commented Jan 23, 2015 at 5:59