Table of Content
The cardinality of a group \(G =\) the order of \(G\).
If \(g \in G\) and for some \(m \in \mathbb{N}\) such that \(g^m = e\), then \(m\) is called exponent of \(g\). The order of \(g\) is the smallest such exponent, denoted by \(\text{ord}(g)\). In this case we say \(g\) is of finite order (otherwise infinite).
As a consequence of Lagrange’s Theorem, the order of an element \(g\) of a group \(G\) divides the group order |\(G\)| in case \(G\) is finite.
\(G = \{e\}\) is unique and \(G \cong \mathbb{Z}_1 = \{\bar 0\}\).
\(G = \{e, a\}\), and \(a^{-1} = a\) (the only choice).
藉由 table（如下）的方式，列出所有 element 經過 binary operation 的結果，再經過比對，可以得知 \(G \cong (\mathbb{Z}_2, +)\)。
| \(\cdot\) | e | a |
|---|---|---|
| e | e | a |
| a | a | e |
| \(+\) | \(\bar 0\) | \(\bar 1\) |
|---|---|---|
| \(\bar 0\) | \(\bar 0\) | \(\bar 1\) |
| \(\bar 1\) | \(\bar 1\) | \(\bar 0\) |
\(\bar 0 \to e, \bar 1 \to a\).
By cancellation law, every column and every row in the table cannot contain a repeated element.
\[G \cong (\mathbb{Z}_3, +)\]只要 \(\vert G\vert\) 是質數，則 \(G \cong \mathbb{Z}_p\) 是唯一可能。
Let \(G = \{e, a, b, c\}.\)
Case \(\rm I\): \(a^2 = b^2 = c^2 = e\)
大家都是自己的 inverse。
透過 table 可以得知
\[G \cong \mathbb{Z}_2 \times \mathbb{Z}_2。\]Case \(\rm II\): \(a^2 = b\)
至少一元素不是自己的 inverse。 可以看出，兩個 case 是所有可能性的一個 partition，其聯集涵蓋所有情況。
一樣透過 table 可以得知
\[G \cong (\mathbb{Z}_4, +)。\]注意！\(G\) 的 order 到 \(4\) 的時候終於出現非唯一的群！
\(G \cong \mathbb{Z}_{6}\) or \(G \cong S_3 \cong D_3\).
如果 \(G\) 可交換，則 \(G\) 和
\[\mathbb{Z}_8, \mathbb{Z}_4\times\mathbb{Z}_2, (\mathbb{Z}_2)^3\]之一 isomorphic；如果不可交換，則和 \(D_4\) 或 \(Q_8\) isomorphic。
\(Q_8 = \{\pm 1, \pm i, \pm j, \pm k \}\).
A multiplicative group is a group whose group operation is identified with multiplication. 但這裡主要討論的是 \(\mathbb{Z}_n\)。
For \(\bar a \in \mathbb{Z}_n\), \(\bar a\) has the multiplicative inverse if and only if \(a\) and \(n\) are coprime.
Proof
First, we show the if part. Consider \(a\) and \(n\) are coprime. By Euclidean algorithm, there exists \((x, y)\) such that
\[ax + ny = \gcd(a, n) = 1.\]We can see that \(\bar x\) is the multiplicative inverse of \(\bar a\) in \(\mathbb{Z}_n\).
Then, we show the only if part. Suppose \(\gcd(a, n) \not = 1\), and \(\bar x\) is the multiplicative inverse of \(\bar a\). Then we can show that
\[ax + ny = 1,\]for some \(y \in \mathbb{Z}\). However, if we divide both sides by \(\gcd(a, n)\),
\[a'x + n'y = {1 \over \gcd(a, n)},\]a contradiction occurs since \(a' = {a \over \gcd(a, n)}, n' = {n \over \gcd(a, n)} \in \mathbb{Z}\), while \({1 \over \gcd(a, n)} \not \in \mathbb{Z}\).
Thus we can conclude that \(a\) and \(n\) must be coprime. ◼
Let \((S, *)\) be an associative binary structure with identity and \(S'\) be a subset of \(S\) consisting all elements with inverse. Then \((S', *)\) is a group.
在「差點形成」（有 identity element 且 \(*\) is associative） group 的 binary structure \((S, *)\) 中，只取出 \(S\) 中有 inverse 的元素成為一 subset \(S'\)，則 \((S', *)\) 為一 group。
Remark
Denote \(\mathbb{Z}_n^{\times}\) the subset of \(\mathbb{Z}_n\) consisting all elements with multiplicative inverse. \((\mathbb{Z}_n^{\times}, \cdot)\) is a group.
Groups 的一大特性就是 symmetry！在正三角形上，有六種操作可以使她不變（三旋轉、三鏡射），相當於 \(G\) such that |\(G\)|\(=6\)。
\(X = \Delta\)（正三角形）, \(Sym(X) = \{e, a, a^2, b, c, d\}\)
\(a\) 是旋轉，\(b, c, d\) 是三種鏡射。
Let \(X\) be a geometric object and \(G\) be the set of symmetries of \(X\). Then it is clear that \(G\) have the following properties.
Therefore, the set of symmetries together with the composition operator form an abstract group.
When \(A\) is just a set, a symmetry on \(A\) is just a bijective function from \(A\) to itself. When \(A\) is finite, its symmetry is also called a permutation of \(A\).