(Non) Faithful Group Action question Consider a regular $9$-gon with symmetry group $D_9$.  There are three separate equilateral triangles that can be constructed using the nine vertices and $D_9$ acts on the set $X$ of the three triangles.  Identify the subgroup that acts trivially.
Attempt:
$D_9$ acts on the set $X$ of three triangles, so $|X| = 3$.  I presume in such a way that the vertices of the triangles are permuted.  $G$ is finite and the action is faithful(This is true I think since it is given initially that the three triangles are constructed using all nine vertices, so any perumtation of the vertices will map a single vertex to another).  Hence we may write $$G \cong \text{subgroup of}\, S_3 = \text{subgroup of}\,S_{|X|}$$  So we want $G \cong \langle \text{id} \rangle \leq S_3$, where id is the identity permutation in $S_3$.  The answer to this question is in fact the subgroup $\left\{e,g^3,g^6\right\} = \langle g^3 \rangle$.  So assign a bijection $e \rightarrow 1\,\,g^3 \rightarrow 2\,\,g^6 \rightarrow 3$ and let $G$ act on itself by the left action $g \cdot h := gh$.
Then when I follow this through, I end up with the permutation $(123) \neq $ id.  Why is my method wrong?
Many thanks  
 A: You identified the subgroup correctly.
Realize $D_{9}$ as the group of permutations on $\Bbb{Z}_{9} = \{ 0, 1, \dots 8 \}$ given by
$$
x \mapsto \pm x + a,
$$
for $a \in \Bbb{Z}_{9}$.
The triangles are $\{0, 3, 6\}$, $\{1, 4, 7\} = \{0, 3, 6\} + 1$, $\{2, 5, 8\} = \{0, 3, 6\} + 2$, so the three cosets of the subgroup $\{0, 3, 6\}$ of $\Bbb{Z}_{9}$.
Clearly the only rotations that fix the triangles are the three elements
$$
\tau_{b} : x \mapsto x + b,
$$
for $b \in \{0, 3, 6\}$.
Now consider an arbitrary reflection
$$
\rho: x \mapsto -x + a.
$$
To fix the triangle $\{0, 3, 6\}$, we must have $a \in \{0, 3, 6\}$. But then $\rho$
this will send $1 \in \{1, 4, 7\}$ to $a - 1 \in \{2, 5, 8\}$, thus swapping the triangles $\{1, 4, 7\}$ and $\{2, 5, 8\}$.
So the subgroup acting trivially on the three triangles is
$$
K = \{ \tau_{0}, \tau_{3}, \tau_{6} \},
$$
and $D_{9}$ induces $S_3 \cong D_3 \cong D_9 / K$ on $X$.
A: A faithful action is one for which the set $\ker * :=\lbrace g | \quad g*x = x \rbrace$ is equal to the set $\lbrace e \rbrace$.
The action of $D_9$ on the set of triangles is not faithful we have to non-identity elements $g^3$ and $g^6$ in the kernel of the action.
You could see that there is a non-identity element in the kernel by noting that associated to every element in $D_9$ there is a corresponding permutation of the set $X$. There $6$ possible permutations of this set but we have $18$ elements in $D_9$. If we think of a mapping from the permutations associated to every element of $D_9$, there are $18$, to one of the $6$ possible permutations then clearly this cannot be a $1-1$ correspondence. Therefore we must have $h_1*x=h_2*x$ for all $x$, with $h_1 \ne h_2$ so $h_2^{-1}h_1 * x = x$. This shows that $h_2^{-1}h_1 \in \ker *$ and $h_2^{-1}h_1 \ne e$. 
So we now know there are at least two elements which act trivially. 
And so the action is not faithful
