Let's say that $D_3$ acts on a bracelet of 3 beads (Denote S), each bead can be Black or White. I want to count the number of different bracelets (4 - I believe)
But using burnside's lemma I get something else... What am I doing wrong?
$R$ := reflection, $S$ := Spin clockwise "vertex to vertex"
$|Orb(S)| = \frac{1}{|D_3|} \sum_{x \in D_3} |Fix(x)| = \frac{1}{6} (|Fix(Id)|+|Fix(R)|+|Fix(S)|+|Fix(S^2)|+|Fix(RS)|+|Fix(RS^2)|) = \frac{1}{6} (8+4+2+2+2+2) \notin \mathbb{Z} $
I have found the problem simple calculation mistake, thanks. $|Fix(RS)| = |Fix(RS^2)| = 4$
non trivial at first were:
$RS(BBW) = R(WBB) = BBW$
$RS^2 (WBB) = RS(BWB) = R(BBW) = WBB$
This is easier (for me, at least) to do in terms of permutations of the vertices, where we have the following fixed point counts: $$123:\ 8$$ $$213:\ 4\mbox{ color scheme AAB}$$ $$132:\ 4\mbox{ color scheme ABB}$$ $$312:\ 2\mbox{ color scheme AAA}$$ $$231:\ 2\mbox{ color scheme AAA}$$ $$321:\ 4\mbox{ color scheme ABA}$$ This gives 4 bracelets.