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


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 $X$.

  • $\begingroup$ Thanks for your response - do you have any comments on my method - I can't see where I made my error. $\endgroup$ – CAF May 2 '13 at 14:21
  • $\begingroup$ @CAF, I believe there is a problem from the start, when you say the action is faithful. It cannot be, because the group $D_9$, of order $18$, is acting on a set $X$ with three elements, so the action is equivalent to a group homomorphism $D_9 \to S_{X} \cong S_{3}$. Since the latter group has order $6$, there has to be a kernel, which ends up being what I called $K$. My approach is essentially to construct explicitly the action of $D_9$ on $X$. $\endgroup$ – Andreas Caranti May 2 '13 at 14:31
  • $\begingroup$ Yes, of course - $D_9$ has order $18$ and $X$ has order $3$, so there is no way for there to be an injection between the two sets, i.e the action is not faithful. So when I said that $D_9$ permutes the vertices, this is not the case here, correct? How else could it act on the triangles? $\endgroup$ – CAF May 2 '13 at 14:43
  • $\begingroup$ Yes, the elements of $D_9$ permute the $9$ vertices of the regular $9$-gon, so you have to check what an element of $D_9$ does to the three triangles, as indicated in my answer. $\endgroup$ – Andreas Caranti May 2 '13 at 14:46
  • $\begingroup$ @BabakS. Thanks!! $\endgroup$ – Andreas Caranti May 5 '13 at 19:25

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

  • $\begingroup$ $g^3$ and $g^6$ both fix each respective triangle within the $9-$gon. Since $G = D_9$ acts on the three triangles by permuting the vertices, the elements $g^3$ and $g^6$ permute the vertices such that the triangles remain in place. Hence they are elements of the kernel of the action. Is this correct analysis? $\endgroup$ – CAF May 2 '13 at 14:59
  • $\begingroup$ Yes though I think it is better to view $D_9$ as permuting the triangles rather than the vertices. Here we are not interested in what happens to vertices. Each symmetry of the 9-gon permutes the triangles in some way. We want to know which of these perform the identity permutation as these will be in $\ker *$ $\endgroup$ – Ben May 2 '13 at 15:12

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