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Apr 13, 2017 at 12:58 history edited CommunityBot
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Aug 10, 2015 at 5:43 comment added Gerhard Paseman If one were able to nearly cover a large S_m by embeddings of a smaller S_{2^n}, that would help give you some idea as to how distributive S_m can be. Even knowing how S_2 embeds in S_5 might help in figuring out f(n) in general. Gerhard "Also Look At Their Clones" Paseman, 2015.08.09
Aug 9, 2015 at 17:55 comment added Joseph Van Name I am familiar with most of the literature on the distributive Laver tables, but there does not appear to be much literature on the "bad" (non-distributive) Laver tables since they do not appear to have much use besides being a construction that includes the distributive Laver tables $S_{2^{n}}$.
Aug 9, 2015 at 17:12 comment added Joseph Van Name If $m,n$ are integers, then the mapping $\phi:S_{2^{n}\cdot m}\rightarrow S_{2^{n}}$ where $\phi(x)=x\,\text{mod}\, 2^{n}$ is a homomorphism, so the algebra $S_{n}$ is not simple whenever $n$ is even. The algebras $S_{n}$ are have no nontrivial automorphisms since $1$ is the unique generator for each of the algebras $S_{n}$. There are many ways to embed Laver tables $S_{2^{n}}$ into larger Laver tables $S_{2^{m}}$ (Drapal has proven some results about embedding smaller Laver tables into larger Laver tables).
Aug 9, 2015 at 3:42 comment added Gerhard Paseman Are these algebras rigid? Primal? Can you "nearly" embed S_n into a larger S_m ? What literature on Laver tables have you not read yet? Gerhard "Always More Questions Than Answers" Paseman, 2015.08.08
Aug 9, 2015 at 3:34 history asked Joseph Van Name CC BY-SA 3.0