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Let $G$ be a (finite) group with subgroup $H$. Pick out (left) coset representatives: $x_1=1, x_2, \dots, x_\ell$ (so $x_1H=1H=H$). Now form an $\ell \times \ell$ table with rows and columns labeled by $H, x_2H, \dots, x_\ell H$.

In addition group cosets together according to their double cosets. For example, if $Hx_2H=x_2H \cup x_3H \cup x_4H$ we group cosets 2, 3, and 4 together, so we would have something like $(H)$ $(x_2H \; x_3H \; x_4H)$ $(x_5H \cdots$ This (parenthesized) list forms the first row of our table.

The rest of the rows are formed by left multiplying by each representative in turn. So, for example, if $Hx_2H=x_2H \cup x_3H \cup x_4H$, we would have

$$\begin{array}{rcccccc} \mbox{Row }x_1H: & \quad & (H) & (x_2H & x_3H & x_4H) & (x_5H & \cdots \\\\ \mbox{Row }x_2H: & \quad & (x_2H) & (x_2x_2H & x_2x_3H & x_2x_4H) & (x_2x_5H & \cdots \\\\ \mbox{etc.} \end{array}$$

Notice that in row $xH$, two cosets $yH$ and $zH$ lie in the same parenthesized group if and only if $x^{-1}yH$ and $x^{-1}zH$ lie in the same double coset. Put another way, cosets $yH$ and $zH$ are connected in row $xH$ if and only if $xHx^{-1}yH = xHx^{-1}zH$. So the parentheses in row $xH$ are recording something about the action of $xHx^{-1}$ on the left cosets of $H$.

Of course if $H$ is a normal subgroup, this is just the Cayley table for the quotient group $G/H$ (with some extra parentheses).

When $H$ is not normal, switching representatives necessarily changes the order in which cosets appear in rows 2+ but the parenthesized groups stay intact. This pseudo-quotient gets as close to a group structure as one can hope for when $H$ fails to be normal.

My question, does this look familiar to anyone? References?

Now for some much needed explanation. About one year ago I got a call out of the blue by someone living close to my university who said he had a short proof of the Feit-Thompson odd order theorem.

I was quite skeptical but told him to email me his proof, I'd look it over and we could talk. It turns out he is a retired mathematician and who has worked off-and-on on his proof for 30+ years as a hobby. While I still have serious doubts his "proof" can be adequately formalized and/or corrected, I did find his fundamental object of study interesting -- these "pseudo-quotients" which he call his "diagrams".

These pseudo-quotients allow one to almost quotient groups even when there are no normal subgroups around. Using these pseudo-quotients, I thought we had a character free proof that the Frobenius Kernel is a normal subgroup, but it fell apart and I haven't been able to repair the proof.

While these quotients haven't helped me prove anything nontrivial (yet), they have caused me to "trip over" a lot of old important theorems.

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Maybe your "pseudo-quotients" are what people call Hecke algebras? See en.wikipedia.org/wiki/Hecke_algebra_of_a_locally_compact_group –  André Henriques Sep 14 '11 at 19:27
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Thanks for the suggestion. I hadn't considered Hecke algebras before. It's definitely worth a look. –  Bill Cook Sep 14 '11 at 20:40
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@Bill: I don't quite understand what to do with your table, rsp. how to interpret it. But somehow your construction reminds me of an association scheme (see en.wikipedia.org/wiki/Association_scheme or the links on blue.utb.edu/zieschang). –  Someone Sep 20 '11 at 11:04
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Absolutely these are examples of "Hecke algebras", which also fit inside the class of "generic algebras" where the structural parameters are allowed to vary outside the cases corresponding to literal groups. –  paul garrett Nov 12 '12 at 21:28
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1 Answer 1

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.

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Thanks Michael! Yes. This does not quite capture the whole picture but is definitely closely related. –  Bill Cook Oct 10 '11 at 11:43
    
Thanks for describing this construction which accounts for all loops! Since I am interested in the sporadic groups and related objects, my question is: how does one obtain the Parker loop of order $2^{13}$ from this? –  DavidLHarden Oct 31 '11 at 22:44
    
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. –  Michael Kinyon Nov 28 '11 at 1:59
    
One does not need to stipulate additionally that every conjugate of $H$ has $T$ as a left transversal. Recall that if $G$ acts on the left cosets of $H$ by left multiplication, the stabilizer of $aH$ in $G$ is the conjugate subgroup $aHa^{-1}$. But the action of $T$ is already sufficient to take any left coset of $H$ to any other, so that remains true with any other left coset of $H$. So it is enough for $T$ to be a left transversal to $H$. –  DavidLHarden Oct 2 '13 at 1:37
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