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Let \(X\) be a \(G\)-set. For any element \(g\in G\), we define
\[X^g = \{x\in X \mid gx=x\},\]which is called the \(g\)-fixed subset of \(X\). Additionally, we define
\[X^G = \bigcap_{g\in G}X^g,\]which is called the \(G\)-fixed subset of \(X\). Note that \(x\in X^G\) iff \(G_x = \{x\}\).
Note that \(x\in X^G\) iff \(G_x = \{x\}\).
為什麼？因為 \(X^G\) 是 \(X^g\) 的聯集，代表 \(X^G\) 含有「對所有 \(g\in G\)，不受 \(g\) 影響的 \(x\)」。
所以說，\(X^g\) 就是那些經過 \(g\) action 也不改變的 \(x\)，而 \(X^G\) 就是經過所有 \(g\) 都不改變的 \(x\)。
另外要注意的是，此處使用的 group action 是「左乘」，但我們也可以設 group action 為 conjugation（group \(G\) acts on \(X\) by conjugation）；也就是說，
\[X^g = \{x\in X \mid gxg^{-1}=x\}\]也是合法的定義。（見 Examples 的最末項。）
最後，\(g\)-fixed subset 可以和 stabilizer 相對應：一個收 \(x\)，另一個收 \(g\)。
Let \(G\) be a group and \(X\) a \(G\)-set. Then we have
\[\sum_{g\in G}\vert X^g\vert = \sum_{x\in X}\vert\text{Stab}_G(x)\vert.\]
Proof
Define a matrix \(M\), whose rows are indexed by \(g\in G\) and columns are indexed by \(x\in X\), with entries
\[m_{g, x} = \begin{cases} 1, &\text{if } gx = x, \\ 0, &\text{otherwise}. \end{cases}\]If we count the number of \(1\)s in \(M\) by the rows, we obtain the left hand side of the equation; on the other hand, if we count it by the columns, we exactly obtain the right hand side. By the double counting principle, the equation is valid. ◼
Let \(p\) be a prime and let \(G\) be a finite \(p\)-group (i.e., the order of \(G\) is a power of \(p\)) that acts on a finite set \(X\). Then it holds that \(\vert X\vert \equiv \vert X^G\vert \pmod p\), where \(X^G\) is the fixed subset of \(X\) under the action of \(G\).
Proof
Let \(X_1,\cdots,X_k\) be the collection of \(G\)-orbits that contain at least two elements. Then we have
\[X = X^G \sqcup X_1 \sqcup \cdots \sqcup X_k.\]By the orbit-stabilizer theorem, for all \(i\), \(\vert X_i\vert\) divides \(\vert G\vert\), implying that \(\vert X_i \vert\) is a multiple of \(p\). Since the \(G\)-orbit form a partition of \(X\), we obtain
\[\vert X\vert = \vert X^G\vert + \vert X_1\vert + \cdots \vert X_k\vert \equiv \vert X^G\vert \pmod p. ◼\]If \(G\) is a finite \(p\)-group for some prime \(p\), then the center \(Z(G)\) is non-trivial. In particular, \(G\) is not a simple group if \(G\) is not of order \(p\) (i.e., \(\vert G\vert = p^2, p^3, \cdots\)).
not a simple group ?
\(Z(G) \triangleleft G\). See Homomorphism & Normal Subgroups.
Proof
Let \(G\) act on \(X=G\) by conjugation. By the theorem of the fixed subset of \(p\)-groups, we have
\[\vert X^G \vert \equiv \vert X\vert \equiv \vert G\vert \equiv 0 \pmod p.\]Note that \(x\in X^G\) iff \(gxg^{-1}=x\) for all \(g\in G\). Thus, \(X^G = Z(G)\), which contains at least the identity. Combining the results, \(Z(G)\) has at least \(p\) elements.
For each element \(x\in G\), the set of all elements conjugate to \(x\) is referred to as the conjugacy class of \(x\), denoted by \([x]\). The stabilizer of \(x\) under conjugation is called the centralizer of \(x\), deonted by \(C_G(x)\). The size of the conjugacy class of \(x\) is \([G:C_G(x)]\), as stated by the orbit-stabilizer theorem.
Let \([G]\) be the collection of all conjugacy classes of \(G\). Then the class equation states that
\[\vert G\vert =\vert Z(G)\vert + \sum_{[x]\in [G], x\not \in Z(G)} [G: C_G(x)].\]Let \(X\) be a \(G\)-set. The centralizer of \(x\in X\) is defined as
\[C_G(x) = \{g\in G \mid gx = xg\}.\]
Under conjugacy, stabilizer becomes centralizer!
針對某一 \(x\)，收集所有使得 \(x\) 可以交換的 \(g\)；也就是說，the centralizer of \(x\) 使 \(x\) 彷彿置身 center！
\(gx = xg,\ \forall g \in C_G(x)\).
Let \(G\) be a group of order \(p^2\), where \(p\) is a prime. Then \(G\) is abelian. Consequently, every group of order \(p^2\) is isomorphic to \(\Bbb Z_p\times \Bbb Z_p\) or \(\Bbb Z_{p^2}\).
Proof
Since \(G\) is a \(p\)-group, by this theorem, \(Z(G)\) is non-trivial. If \(Z(G) = G\), \(G\) is abelian. Suppose that \(Z(G)\) is a proper subgroup of \(G\). Then \(Z(G)\) and \(G/Z(G)\) are both cyclic groups of order \(p\). Let \(Z(G) = \langle a \rangle\) and \(G/Z(G) = \langle bZ(G)\rangle\) for some \(b\not \in Z(G)\). Then every element in \(G\) is of the form \(b^ia^j\). On the other hand, this implies that
\[b(b^ia^j) = b^i(ba^j) = (b^ia^j)b,\]since \(a\in Z(G)\). Thus we can conclude that \(b\) is also contained in \(Z(G)\), which is a contradiction. ◼
Let \(X\) be a \(G\)-set. A subset \(Y\) of \(X\) is called \(G\)-invariant if for all \(g\in G\), \(gY \subset Y\).
In particular, \(Y\) is also a \(G\)-set.
To prove this, we have to show that
Since for all \(g\in G\), \(gY\subset Y\), (1.) is satisfied. (2.) and (3.) are trivially satisfied since \(Y\) is a subset of a \(X\), which is a \(G\)-set. ◼
Suppose that the order of a group \(G\) is divisible by a prime \(p\). Then \(G\) has an element of order \(p\).
Suppose that the order of \(G\) is divisible by a prime \(p\). Consider the \(p\)-th power of \(G\),
\[G^p = \{(g_1, \cdots, g_p)\mid g_i \in G \},\]and let \(\sigma = (1, \cdots, p) \in S_p\). The group \(P = \langle \sigma \rangle\) acts on \(G^p\) by permuting the elements of \((g_1, \cdots, g_p)\).
Let \(Y = \{(g_1, \cdots, g_p)\in G^p \mid g_1\cdots g_p=e \}\), where \(e\) is the identity element of \(G\). We observe that:
By this theorem, we have
\[\vert Y^P\vert \equiv \vert Y\vert = \vert G\vert^{p-1}\equiv 0 \pmod p.\]Note that \(y\in Y^P\) iff \(y = (g,g,\cdots,g)\) for some \(g\in G\) with \(g^p = e\), since \(Y^P \subset Y\). Because \(\vert Y^P\vert \equiv 0\pmod p\) and \((e, e, \cdots, e) \in Y^P\), \(Y^P\) must contain at least \(p\) elements. Then for any \((g,\cdots, g)\in Y^P\) with \(g\not = e\), the order of \(g\) is \(p\), which satisfies the conclusion of this theorem. ◼
只有形如 \((g, g, \cdots, g)\) 這樣的 element 才不會被任何 \(\sigma_i \in P\) 影響！
Lagrange’s theorem 指出，若有 order 為 \(n\) 的 element，則 group size 可被 \(n\) 整除。反過來成立嗎？若 \(G\) 是交換群，根據 the fundamental theorem of finitely generated abelian groups，\(G\) 可以被拆為數個 \(\Bbb Z_{p_i^{e_i}}\) 的 direct product，其中 \(p_i\) 是 \(\vert G \vert\) 的質因數且 \(e_i \in \Bbb N\)。可以觀察到，\(\Bbb Z_{p^k} = \langle 1 \rangle\) 所有元素的 order 涵括了 \(p^k\) 的所有因數：\(\langle p^i\rangle\) 的 order 正是 \(p^{k-i}\)。於是對於 abelian group，Lagrange’s theorem 的逆定理是成立的。
但若不是交換群呢？柯西定理就派上用場了，只不過多了質數的限制。