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\(S_G\): \(G\) 的 permutation group。
We have \(S_G \cong S_{\vert G\vert }\).
Every group is isomorphic to a group of permutation. More percisely, every group \(G\) is isomorphic to a subgroup of \(S_G\) given by:
\[\begin{align*} \lambda_a &: G \to G, \text{ defined by } \lambda_a(g) = ag, \tag{1} \\ \phi &: G \to S_G, \text{ defined by } \phi(g) = \lambda_g, \tag{2} \end{align*}\]for all \(a, g \in G\). Since \(\phi\) is injective, \(G\) must be isomorphic to a subgroup of \(S_G\).
Note:
We can also say that the mapping is given by
\[\phi: a \mapsto \lambda_a,\]where \(\lambda_a(g) = ag\).
\(\to\) 和 \(\mapsto\) 是不一樣的！\(f: A\to B\) 代表 \(A\) 是 function 的 domain 而 \(B\) 是 codomain，用來定義 function；\(f: x \mapsto y\) 說明了 \(f\) 究竟如何 map input (\(x\)) to output (\(y\))，例如 \(f: x \mapsto x^2\) 相當於 \(f(x) = x^2\)。
Remark
\(\lambda_a\) 在 \((1)\) 作為一個 mapping，其實也是一 permutation（底下會證明 \(\lambda_a\) 是 permutation）。permutation 本來就是一 image 和 preimage 相同的 one-to-one、onto mapping！（想想 permutation 總是以 composition 進行運算。）
見 Automorphism。
Let \(G=\mathbb{Z}^{\times}_8 = \{\bar 1, \bar 3, \bar 5, \bar 7 \}.\) Then
\[\lambda_{\bar 3} = \begin{pmatrix} \bar 1 & \bar 3 & \bar 5 & \bar 7 \\ \bar 3 & \bar 1 & \bar 7 & \bar 5 \end{pmatrix}.\]全部元素乘上 \(\bar 3\)。 \(\lambda_{\bar 3}(\bar 5) = \bar 3\cdot\bar 5 = \bar 7\)。
Cayley’s theorem states that \(G\cong \{\lambda_{\bar 1},\lambda_{\bar 3},\lambda_{\bar 5},\lambda_{\bar 7}\} \le S_G\).
To prove this theorem, we first claim that
Suppose the above are true. Let \(\hat G = \{\lambda_a:a\in G \}\). We then claim that
Proof of claim 1.
We need to show that for each \(a \in G\), \(\lambda_a\) is one-to-one and onto.
For all \(g,h \in G\), if \(\lambda_a(g) = \lambda_a(h)\), then \(ag = ah\). By the cancellation law, we have \(g = h\).
For all \(h \in G\), let \(g=a^{-1}h\). Then \(\lambda_a(g) = a(a^{-1}h) = h\).
Proof of claim 2.
\(\lambda_a \circ \lambda_b(g) = \lambda_a(\lambda_b(g)) = \lambda_a(bg) = a(bg)\). By associativity, this is equal to \((ab)g = \lambda_{ab}(g)\).
代 \(g\) 進來才好證。
Proof of claim 3.
To prove this, the following three conditions must be satisfied.
Since for all \(a, b\in G\), we have \(ab \in G\); this is proved by claim 2.
Since \(\lambda_e(g) = g\) for all \(g \in G\), \(\lambda_e\) is the identity of \(S_G\), which is contained in \(\hat G\).
By claim 2, we have \(\lambda_{g^{-1}} \circ \lambda_g = \lambda_{g^{-1}g} = \lambda_e\) and likewise \(\lambda_{g} \circ \lambda_{g^{-1}} = \lambda_{gg^{-1}} = \lambda_e\). Thus \(\lambda_{g^{-1}}\) is the inverse of \(\lambda_g\), which is contained in \(\hat G\).
Proof of claim 4.
We need to verify that the function \(\phi:G \to \hat G\) defined by \(\phi(a) = \lambda_a\) satisfies three coonditions below:
For all \(a,b\in G\), if \(\phi(a)=\phi(b)\), we have \(\lambda_a = \lambda_b\). Then \(\lambda_a(g) = \lambda_b(g)\) for all \(g \in G\); in particular, \(\lambda_a(e) = \lambda_b(e)\). From this we can conclude that \(a=b\).
For all \(\lambda_a \in \hat G\), we have \(a \in G\) such that \(\phi(a) = \lambda_a\).
We need to show that \(\phi(ab) = \phi(a) \circ \phi(b)\), that is \(\lambda_{ab} = \lambda_a \circ \lambda_b\), which is already proved in claim 2. ◼
若要研究 group \(G\) such that
\[|G|=n,\]我們可以只針對 \(S_n\) 的 subgroups 來縮小研究範圍。又已知 the order of an element of \(G\) divides |\(G\)|，於是我們可以從找出 order 為 \(d,\ d \vert n\) 的 element 開始，進而找出所有 subgroups of order \(n\)。