# Special Isomorphism?

Visualise your right hand in front of you, and put your thumb, second and third finger in a configuration such that they are all perpendicular to one another(like the right hand rule). Draw 3d axes, and allow the hand to rotate such that under any rotation, you must have the fingers lying on the axes(imagine you can keep on rotating your wrist infinitely).

Clearly , there are 24 different positions possible. My question is , firstly, do the rotations form a group, which I assume they do , and if so, what is this group isomorphic to? My guess would be a permutation group of some kind , but I would appreciate some feedback on this.

• "there are 24 positions possible" are you sure? If I understand you correctly you're counting 3-element subsets of $\{e_1, e_2, e_3, -e_1, -e_2, -e_3\}$ of the same orientation as the standard basis $e_1, e_2, e_3$; there are $\binom{6}{3} = 20$ candidates of which exactly half have the correct orientation, for a total of 10 possible positions. Commented Feb 11 at 23:08
• My logic was that you can place the thumb on the axes in 6 positions, then the second finger can be placed in 4 positions for each of those 6 positions, and then the third finger is fixed due to required orthogonality of the third finger in a right-handed manner.
– J.D
Commented Feb 11 at 23:10
• ...right, disregard my nonsense above :) Commented Feb 11 at 23:16
• All good , I appreciate the interaction with my post regardless :)!
– J.D
Commented Feb 11 at 23:17
• I think generalized to dimension $n$ the group could be isomorphic to a index $2$ subgroup of $\mathbb{Z}^n_2 \rtimes S_n$, where $S_n$ acts by the natural action on $\mathbb{Z}_2$. Commented Feb 11 at 23:29

Let $$\mathcal{B}=(e_1,...,e_n)$$ be the standard basis of $$\mathbb{R}^n$$. Let $$A = \mathbb{R}e_1 \cup ... \cup \mathbb{R}e_n$$ be the axes. Then we need all matrices $$R \in SO(n)$$ with $$R\mathcal{B} \subseteq A$$, i.e. if $$R = (r_1,...,r_n)$$ with $$r_i \in \mathbb{R}^n$$ it has to be $$r_i = \lambda_i e_i, \ i = 1,...,n$$ for some $$\lambda_i \in \mathbb{R}$$. Since $$R$$ is orthogonal and has determinant $$1$$, the matrix $$R$$ is of the form $$R = (\varepsilon_1 e_{\sigma(1)},...,\varepsilon_n e_{\sigma(n)})$$ with $$\varepsilon_i = \pm 1, \sigma \in S_n, \det R = 1$$. So, $$R = \text{diag}(\varepsilon_1,...,\varepsilon_n)\cdot (e_{\sigma(1)},...,e_{\sigma(n)})$$ ($$\text{diag}$$ diagonal matrix), and the group of all these matrices is $$G = \{R \in G' \ | \ \det R = 1\}$$ with $$G' = \{\text{diag}(\varepsilon_1,...,\varepsilon_n)\cdot (e_{\sigma(1)},...,e_{\sigma(n)}) \ | \varepsilon_i = \pm 1, \sigma \in S_n\} \subseteq O(n)$$.

Now, $$G' \cong (\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$$, and the index $$[G':G] = 2$$. $$G'$$ has $$2^n \cdot n!$$ elements and therefore $$G$$ has $$2^{n-1}n!$$ elements. For $$n = 3$$ $$G$$ has $$2^2\cdot 3! = 4 \cdot 6 = 24$$ elements.