User michael kinyon - MathOverflow

Answer by Michael Kinyon for Parker-like loop of order 2187?

2011-11-28T02:05:12Z

To follow up on Henry Cohn's answer, see also:

A. Drapal and P. Vojtechovsky, Code loops in both parities, Journal of Algebraic Combinatorics 31 (2010), no. 4, 585-611.

Answer by Michael Kinyon for (Dis)similarity between groups and Lie algebras

2011-10-11T02:25:19Z

This isn't getting directly at your question, but you might consider looking at Ellis' notion of a multiplicative Lie algebra. This all started in:

G.J. Ellis, On ﬁve well-known commutator identities, J. Aust. Math. Soc. Ser. A 54 (1993), 1–19.

Then just search for "multiplicative Lie algebra" or "multiplicative Lie ring" for more recent papers. Multiplicative Lie rings provide an interesting framework in which to think about some group and Lie algebra results in a unified way.

Answer by Michael Kinyon for Has this "pseudo-quotient" of groups been studied before?

2011-10-10T02:24:57Z

This is very interesting. I don't really understand what is going on with the double cosets, but this is reminiscent of the standard construction of loops (quasigroups with identity elements) as tranversals in groups. This dates back to Baer in the early 40's. The idea is the following. You have a group $G$, a subgroup $H$, and a transveral $T$, that is a set of coset representatives. Assume further that $T$ is normalized, that is, $1\in T$. For $x,y \in T$, define an operation $\circ$ on $T$ by $(x\circ y)H = xyH$, that is, $x\circ y$ is the unique representative in $T$ of $xyH$. It is easy to see that $1$ is the identity element for $\circ$ and that for all $a,b\in T$, the equation $a\circ x = b$ is uniquely solvable for $x\in T$. If, in addition, one has that $T$ is a transversal for every conjugate of $H$, then it follows that the equation $y\circ a = b$ is also uniquely solvable for $y\in T$. Thus $(T,\circ)$ is a loop. Conversely, it is not hard to show that every loop arises in this way.

The tables you are talking about remind me of this, except that you are keeping track of double coset clusters and you are not worrying about which products of representatives themselves have representatives. It would be interesting to know how this idea relates to loop theory.

Comment by Michael Kinyon

2011-11-28T01:59:21Z

David, I'm not an expert in code loops and don't know offhand what the multiplication group of the Parker loop is. (The multiplication group of a loop is the group generated by the loop's left and right translations.) I would suggest looking more generally at the literature on code loops, especially the work of my colleague Petr Vojtechovsky. He can probably give you more details on the particulars of the Parker loop.