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My task: Describe all finite subgroups of SO(3) that contain the elements $$ H:= \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \qquad \text{and} \qquad G:= \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix}. $$

What I've got so far: We have 5 isomorphtypes:$$ C_n, D_{2n},A_4,S_4, A_5$$ Now I now the subgroup must atleast be of order 4 elements because of: $$\langle G,H \rangle=(h,g,hg,e)$$ Also I can not find a subgroup in $$C_n$$ because I have 2 generators with 2 separate cycles so $$C_n$$ is out of the question. But now I'm stuck. I tried to calculate the stabilizer but they are always of order 1 so the orbit is as big as the subgroup itself and I also do not know exactly, how big my subgroup is (only know that it is atleast of order 4), so I can not use Lagrange's Theorem. I tried working with Sylow but I'm stuck here. Thank you for your help

  • What is the question exactly? You want to show which groups on your list can be realized by a subgroup of $SO(3)$ that contains $G$ and $H$? – Travis Willse Nov 23 '14 at 10:41
  • Hello Travis, yes exactly. I should describe all subgroups of SO(3) who contain G and H. –  Nov 23 '14 at 11:03
  • Do you have a visual intuition for how each of the groups is realized in $SO(3)$? – Travis Willse Nov 23 '14 at 11:10
  • Well I know that A_4 is isomorphic to the tetrahedral rotation group, S_4 to octahedral rotation group and A_5 icosahedral rotation group. And dihedral is generated by 2Pi/n and a rotation by pi and cyclis y rotation by 2pi/n. But I honestly have troubles understanding this and clue how this is helpful here –  Nov 23 '14 at 12:20
  • I've written up a worked solution for the $S_4$ case as an answer below. The other cases are similar (if sometimes more subtle). – Travis Willse Nov 23 '14 at 12:40
  • Thank you so much for your help Travis. But I don't understand why G and H are reflections and not rotations? –  Nov 23 '14 at 12:46
  • They are each reflections through a line, which is the same as a rotation by $\pi$ about that line. And you're welcome, I hope you found it useful. – Travis Willse Nov 23 '14 at 12:49
  • I have a question: Why a finite subgroup in $SO(3)$ must be of the following types? Thanks. –  Nov 23 '14 at 12:53
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    @John: The proof is relatively short, and interestingly amounts to finding the unordered triples $(a, b, c)$ of positive integers such that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} > 1$. See: http://groupprops.subwiki.org/wiki/Classification_of_finite_subgroups_of_SO%283,R%29 – Travis Willse Nov 23 '14 at 13:29
  • @Travis: Thanks for the information! –  Nov 23 '14 at 13:32
  • You're welcome, I hope you found it useful! – Travis Willse Nov 23 '14 at 13:45

1 Answers1

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As you've implied, $G$ and $H$ are commuting reflections through lines (which must hence be identical or orthogonal), so the question is really just asking which of the groups in your list admits two such reflections.

Like you say, $S_4$ can be realized as the group of orientation-preserving symmetries of the octahedron, which has two such symmetries. In fact, if we take the octahedron to be the one with vertices $(\pm 1, 0, 0)$, $(0, \pm 1, 0)$, and $(0, 0, \pm 1)$, then direct checking (staring at the involved matrices and vectors, really) shows the matrices $G$ and $H$ are two such transformations.

Travis Willse
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