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I am working with the set $X = \mathbb{Z}_{12}$. On this set I have defined two functions $T:X \rightarrow X, \ T(x)=x+1$ and $I:X \rightarrow X,\ I(x)=-x$. It is known that $G:= \left \langle T,I \right \rangle \cong D_{12}$.

$G$ works on the set consisting of all subsets of $X$ with $6$ elements. I call this set $\mathcal{X}_6$. So, $\mathcal{X}_6 = \left \{ A \subset X \ | |A|=6 \right \}$. I believe a way to define the action is $g(A) = \left \{ g(x_1),...,g(x_n) \right \}$. I have to show that the action of $G$ on $\mathcal{X}_6$ has $50$ orbits.

To prove this I must use the counting theorem. I have come this far: for the identiy element we have $\mbox{id}(A)=A$ for all $A \in \mathcal{X}_6$ and there are $\binom{12}{6}$ such sets $A$. So $|\mathcal{X}_6^{id}|= 924$. For all $T^n$ with $n \not = 0$ I believe there are no $x$ such that $x+n=x$ therefore I conclude

$ |\mathcal{X}_6^{T^n}| = 0 \ \ \mbox{for all $n=1,...,11$} $

Then for an element of the form $IT^n(x) = -x-n = x$ there are two $x,y \in X$ if $n$ is even. For example if $n=0$ then we have $x=0$ or $x=6$. However, now we cannot find a subset with $6$ elements, therefore I think

$ |\mathcal{X}_6^{IT^n}| = 0 \ \ \mbox{for all $n=0,1,...,11$} $

However, If I use the counting formula I do not get the correct answer. Can someone please explain where my reasoning is wrong? Much thanks in advance!

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    $\begingroup$ You have accidentally use $I$ for the identity and one of your maps. Also a hint for where you went wrong - let $A=\{1,2,3,11,10,9\}$, then $I(A)=A$. $\endgroup$ – Robert Chamberlain Oct 23 '17 at 9:24
  • $\begingroup$ Thank you very much for your comment and hint! It is greatly appreciated. I will fix the identity. $\endgroup$ – user421927 Oct 23 '17 at 12:03
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I would propose, rather than looking at it arithmetically, the geometrical insight:

one can imagine the problem as symmetries of regular dodecahedron, where $T^n$ corresponds to rotation about $\pi /6$ and $R_i := IT^n$ reflections with axis passing either through vertices ($n$ even) or midpoints of edges ($n$ odd).

In the latter case, each of the elements $X_6^g$ is easily described by the triple of points on one side of the axis (plus their images), hence $\begin{pmatrix}6 \\ 3 \end{pmatrix}$ elements.

But notice that rotations $T^2$ and ${T^{10}}$ has as its $X^{T^2} = X^{T^{10}} = \{{odd \,numbers}, {even \,numbers}\}$, $X^{T^6}$ consists of triples of couples of points opposite to each other, hence $\begin{pmatrix}6 \\ 3 \end{pmatrix}$ elements each and $T^{4}$, $T^{8}$ have their corresponding sets equivalent to couples of triplets of vertices (e.g. $T^{4}\{0, 4, 8\} = \{0, 4, 8\} $), $\begin{pmatrix}4 \\ 2 \end{pmatrix}$ elements each.

In total this really makes 1200.

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  • $\begingroup$ Thank you very much for your very clear explanation! I $\endgroup$ – user421927 Oct 23 '17 at 14:43
  • $\begingroup$ Could you please explain where you get the 6 and 3 for the elements $X^g_6$? $\endgroup$ – user423841 Oct 24 '17 at 12:19
  • $\begingroup$ @user423841 There are six points on each side of the axis in case it passes through the midpoint and any set fixed by a reflexion $g$ is described by three points on one side (the other three are their (pre)-images under $g$). There are $\begin{pmatrix} 6 \\ 3 \end{pmatrix}$ of such triplets. In case the axis passes through opposite vertices the fixed sets either consist of three vertices (out of five) non lying on the axis - $\begin{pmatrix} 5 \\ 3 \end{pmatrix}$ possibilities - or out of two plus the two vertices on the axis, hence $\begin{pmatrix} 5 \\ 2 \end{pmatrix}$ possibilities. $\endgroup$ – pepa.dvorak Oct 26 '17 at 16:17

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