4
$\begingroup$

Let $X$ be the set of $2 \times 2$ matrices over field $\Bbb F_2$, and $G \subset X$ be the group of invertible $2 \times 2$ matrices over $\Bbb F_2$

Let $G$ act on $X$ by $g*x = gxg^{-1}$

Need to find the orbits of $X$ under this action.

Computing for each $x \in X$ would require $6 \times 16 =96$ calculations...What is a quicker way to obtain the distinct orbits?

$\endgroup$
  • 3
    $\begingroup$ What are some of your observations? Are there special cases that make the partitioning of $X$ into orbits easy? Are there some properties that all matrices in an orbit must share? Partitioning the sixteen matrices into orbits should go pretty quickly once you find a few orbits; those matrices are taken into account and don't need to be computed again. $\endgroup$ – hardmath Mar 2 '15 at 13:53
  • 3
    $\begingroup$ @hardmath ok, thanks for reminding me that the orbits partition $X$ :) (I should know this if I had read my notes more carefully! and the fact that the orbits either do not intersect or are identical). So once I've noticed that the orbits I've computed cover $X$ I can stop computing lol. This shouldn't take long as you said $\endgroup$ – AMFS Mar 2 '15 at 14:05
  • 1
    $\begingroup$ Note that the group action (similarity) involves $|G|=6$, so sizes of orbits have to divide six. There are a couple of obvious singleton orbits, and similarity preserves rank. So you might want to start with a random matrix in $X$ that has rank one, just to see how big its orbit is. Finally, transpose will either give the same orbit or a "mirror image". $\endgroup$ – hardmath Mar 2 '15 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.