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Is there any strategy to find out the number of symmetries of any geometric figure?

Can you please explain applying to the following figure where all triangles are equilateral: figure

By exhaustion, I have found 6 rotatations and 6 reflections wich sum up to 12 symmetries. Is this correct?

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    $\begingroup$ It belongs to $D_{3h}$ group. $\endgroup$ – Ng Chung Tak Apr 4 '17 at 8:54
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In general, when you need to find the symmetry group of a figure, the best way to deal with the question is to find a nice set on which this isometry group acts. For instance :

  • the isometry group of the tetrahedron acts on the four vertices of the tetrahedron. This eventually leads to the isomorphism between the isometry group of the tetrahedron with $S_4$ (the symmetric group over four elements).

  • the isometry group of the cube acts on the four big diagonals of the cube. This eventually leads to the isomorphism between the direct isometry group of the cube with $S_4$.

Let $X$ be your figure $O$ be its center and $G$ be its isometry group. Clearly, $G$ acts on the faces of $X$, furthermore a square is sent by $G$ on a square. As a result, denoting $F_1=(2,3,6,5)$, $F_2=(2,4,7,5)$ and $F_3=(3,4,7,6)$, the group $G$ acts on $\mathcal{F}:=\{F_1,F_2,F_3\}$ and therefore you have a group morphism from $G$ to $S_{\mathcal{F}}$.

I claim that this group morphism is onto. Furthermore, I claim that the kernel of this group morphism is of cardinal $2$ (to do this, you can remark that if $C_i$ the center of the face $F_i$ then if $g$ is in the kernel then $g\cdot OC_i=OC_i$ for all $i=1,2,3$).

Therefore you have $2\times |S_3|=12$ elements in your group.

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