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Let \(G\) be the group of symmetries of some object \(X\). Fix a point \(x\) in \(X\), let
\[G_x = \{gx|g \in G\},\]called the \(G\)-orbit of \(x\). Suppose \(G_x\) is finite and write
\[G_x = \{x_1=x, x2,\cdots,x_n\} = \{g_1x,g_2x,\cdots,g_nx\}\]for some \(g_i \in G\). Set
\[G_i = \{g \in G | gx=x_i\}.\]It is clear that \(G\) is the disjoint union of all \(G_i\). Moreover,
\[G_1 = \{g \in G | gx=x\}\]is also called the stabilizer of \(x\) in \(G\), denoted by \(\text{Stab}_G(x)\), which is a subgroup of \(G\).
\(G_x\)：\(x\) 經過 \(g\) 變換可以變成的 object set；不論有多少 \(g\) 使得 \(gx = x_i\)，\(x_i\) 在 \(G_x\) 只 count \(1\)！
\(G_i\) 只取可使 \(x\) 變換成 \(x_i\) 的 symmetry set。
Stabilizer 收集所有使 \(x\) 保持不變的 symmetry。
\(g_iG_1 = G_i\) for all \(i\).
Proof
Consider the map \(\phi_i: G_1 \to G_i\), given by \(\phi_i(g) = g_ig\). We have for all \(g \in G_1\), \(\phi_i(g)x = g_igx = g_ix = x_i\). Therefore, \(\phi_i(g) \in G_i\).
On the other hand, let \(\rho_i: G_i \to G_1\) given by \(\rho_i(g) = g_i^{-1}g\). Since both \(\rho_i \circ \phi_i\) and \(\phi_i \circ \rho_i\) are equal to the identity map, \(\rho_i\) is the inverse map of \(\phi_i\).
Hence, \(\phi_i\) is a bijection, which implies that
\[|G_1| = |G_i|\]and \(G_i = g_iG_1\). ◼
identity map: \(f(x) = x\).
Let \(G\) be a group of symmetries of \(X\). Fix \(x \in X\) and let \(G_x = \{g_1x,\cdots,g_nx\}\). Then
\[|G| = |G_x| \dot{} |\text{Stab}_G(x)|\]and
\[G = \bigsqcup_{i=1}^{n}g_i\text{Stab}_G(x).\]
正多邊形
Let \(X\) be a regular \(n\)-gon with the group of symmetries \(G\). Let \(o\) be the center of \(X\) and \(x\) be one of its vertex. We first have that \(G_x\) is equal to the set of \(n\) vertices of \(X\). Moreover,
\[G_x = \{x, \sigma(x),\cdots,\sigma^{n-1}(x)\},\]where \(\sigma\) is the rotation around the center \(o\) of \(2\pi/n\) degree.
在 symmetry 下，頂點只能變換到另一頂點，所以 \(G_x\) is equal to the set of \(n\) vertices of \(X\).
And the elements in the stabilizer \(\text{Stab}_G(x)\) must fix the line \(\overline{ox}\). Thus, \(\text{Stab}_G(x) = \{e,\tau \}\), where \(\tau\) is the refection through \(\overline{ox}\).
Thus we have
\[|G| = |G_x||\text{Stab}_G(x)| = 2n\]and
\[G = \bigcup^n_{i=0}\sigma^i\{e,\tau\} = \{e,\sigma,\cdots,\sigma^{n-1},\tau,\sigma\tau,\cdots,\sigma^{n-1}\tau\}.\]Finally, we have \(\sigma \tau = \tau \sigma^{-1}\). Especially, \(G\) is not abelian.
因為鏡射 + 旋轉依然是鏡射，所以
\[\tau' = \sigma \tau = (\tau')^{-1} = \tau^{-1}\sigma^{-1} = \tau \sigma^{-1}.\]
The group of a regular \(n\)-gon is called the dihedral group, denoted by \(D_n\), which is a group generated by \(\tau\) and \(\sigma\) satisfiying the following conditions:
\[\begin{eqnarray} \sigma^n = \tau^2 = e, \tag{1} \\ \sigma \tau = \tau\sigma^{-1} \tag{2}. \end{eqnarray}\]\((2)\) implies that every element in the group can be written as the form
\[\sigma^i \tau^j.\]Therefore, if a group satisfies the above conditions, it has at most \(2n\) elements. \((2)\) also uniquely determiines a binary structure, i.e. the above conditions uniquely determined the group structure of \(D_n\), and we use the following notation as well:
\[D_n = \langle \sigma, \tau \vert \sigma^n = \tau^2 = e,\ \sigma \tau = \tau \sigma^{-1} \rangle,\]which defines a group by generators and relations.
\(G_n = \langle \text{generator} \vert \text{relations} \rangle\).