NB: For this question, everything is finite.

Recently I've been fascinated by the following two observations:

  • Almost all loops have a trivial automorphism group (McKay & Wanless, 2005, in the context of Latin squares).
  • Almost all groups have a non-trivial automorphism group (in fact, all groups of order $n \geq 3$ admit a non-trivial automorphism).

However, the only difference between loops and groups is "cancellation" vs. "associativity". (Actually, in a loop, an element's left inverse might not equal its right inverse, but the equality of left and right inverses in a group does not need to be included as a group axiom.)

Question: How can we strengthen the loop axioms so as to preserve the property that "almost all X have a trivial automorphism group"?

Question: How can we weaken the group axioms so as to preserve the property that "almost all Y have a non-trivial automorphism group"?

Question: Is the a set of axioms "between" the loop axioms and the group axioms for which "almost all Z have a trivial automorphism group" and "almost all Z have a non-trivial automorphism group" are both false?

  • $\begingroup$ mathworld.wolfram.com/AlgebraLoop.html $\endgroup$ – Lee Mosher Jul 14 '12 at 14:55
  • $\begingroup$ Most semigroups seem to have trivial automorphism (at least that is the result of an enumeration of small semigroups -- see the thesis of Andreas Distler, available on the internet.) A similar question (automorphism groups of free monoids vs. free groups) is at math.stackexchange.com/questions/109600. Note that semigroups and monoids obey the associativite law and yet seem to be non-symmetric in the sense that many have nontrivial automorphism groups. $\endgroup$ – Ken W. Smith Jul 14 '12 at 15:55

I suggest a general algebraic approach for this problem, since to me the major thing that is changing is the idea of 'many". Although you haven't said so, you seem to be asking about a demarcation in the lattice of varieties of algebra with one binary operation where one side has algebras with more than one automorphism versus those that have only one. The latter are called rigid, and since no nontrivial variety has only rigid algebras (think of powers), you will need to have a good technical definition of 'many'. Perhaps pseudovarieties are the classes of interest.

I recommend looking at studies of rigid algebras. If you are interested in equational formulations which promote rigidity, you could do worse than looking at versions of primal algebras, which are very rigid. The varieties they generate are called arithmetical, and therein might lie part of the answer you seek.

Gerhard "Ask Me About System Design" Paseman, 2012.07.14


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.