Timeline for Has this "pseudo-quotient" of groups been studied before?
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| Oct 2, 2013 at 1:37 | comment | added | DavidLHarden | 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 $H$. | |
| Nov 28, 2011 at 1:59 | comment | added | Michael Kinyon | 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. | |
| Oct 31, 2011 at 22:44 | comment | added | DavidLHarden | 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? | |
| Oct 10, 2011 at 11:43 | comment | added | Bill Cook | Thanks Michael! Yes. This does not quite capture the whole picture but is definitely closely related. | |
| Oct 10, 2011 at 2:24 | history | answered | Michael Kinyon | CC BY-SA 3.0 |