0

Let $G \subseteq S_8$ be the subgroup generated by some 4-cycles. If we number the elements $1,2,\dots, 8$, the 4-cycles are

  • $(1234),(5678),(1485),(2376),(1265),(4378)$

I am not sure if I have the list write... I am thinking of the corners of a cube and rotating them. Let me just write the vertices as elements of $\mathbb{Z}_2^3$. What is the order of this group?

These generators are twists of the cube:

  • $\boxed{000 \to 001 \to 011 \to 010}$ $\boxed{100 \to 101 \to 111 \to 110}$
  • cyclic permutations of these two by $abc \mapsto bca$

What are the relations among these generators to give a presentation? In the case of the dihedral group we get some relations like:

$$ D_{2n}= \langle r,s \mid r^{n}=s^{2}=1, rs=sr^{-1} \rangle $$

I am looking something similar for my $G$ generated by 4-cycles. Or can this group be an action on a vector space? Right now the twist only works on half the elements.

EDIT The comments suggest $\mathbf{G = S_8}$. Can someone find me a way to get a 2-cycle $(12)$ from these 6 elements?

Community
  • 1
cactus314
  • 24,438
  • There are ${8\choose 4}\cdot 3!$ 4-cycles. That is much more than the 6 you've written. – tomasz Jun 25 '14 at 15:06
  • @tomasz I am asking about the subgroup generated by the six 4-cycles I have listed. – cactus314 Jun 25 '14 at 15:07
  • You should say then that $G$ is the group generated by these cycles. Right now it looks like you're saying that $G$ is generated by all 4-cycles and you're attempting to write them all. – tomasz Jun 25 '14 at 15:08
  • Also, this is certainly not a subgroup of the isometry group of a cube. Any isometry of a cube that fixes $3$ points is constant, while all the 4-cycles will inevitably have $4\geq 3$ fixed points. If anything, you're rotating walls, like $2\times 2\times 2$ Rubik's cube. – tomasz Jun 25 '14 at 15:09
  • Those six $4$-cycles generate $S_8$. – Derek Holt Jun 25 '14 at 15:13
  • @tomasz group $G$ might have an action on some other vector space – cactus314 Jun 25 '14 at 15:15
  • @johnmangual: Certainly. Any finite group admits a finite dimensional, faithful representation. But the natural action on ${\bf Z}_2^3$ is certainly not linear, since it has no fixed points. – tomasz Jun 25 '14 at 15:19
  • @DerekHolt can you find me a word in these generators giving the transposition $(12)$ ? – cactus314 Jun 25 '14 at 15:27
  • If you call the six $4$-cycles $a,b,c,d,e,f$, then $(1,2) = a c b^{-1} d^{-1} e^{-1}$ (where permutations are composed from left to right). – Derek Holt Jun 25 '14 at 22:37
  • @DerekHolt Thanks for your reply. My main question is about the Cayley graph generated by these 4-cycles, but for today I settled on this easier question I could state more precisely. – cactus314 Jun 26 '14 at 02:47

0 Answers0