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Let \(G\) be a group and let \(a \in G\). Then
\[H = \{a^n | n \in \mathbb{Z}\}\]is a subgroup of \(G\) and it is the smallest subgroup of \(G\) that contains \(a\), i.e. every subgroup containing \(a\) contains \(H\).
identity: \(a^0 = e\); inverse: \(a^{-n}a^n = e\)
為什麼是 smallest？因為封閉性。
Let \(G\) be a group and \(a \in G\). Then the subgroup \(\{a^n:n\in \mathbb{Z}\}\) is called the cyclic subgroup generated by \(a\), and denoted by \(\langle a \rangle\).
A group \(G\) is cyclic if \(G = \langle a \rangle\) for some element \(a \in G\). The element \(a\) is called a generator for \(G\).
Generator for a cyclic group \(G\) is not necessarily unique. For example, under addition, \(\mathbb{Z}_n = \langle \bar 1 \rangle = \langle \overline{n-1} \rangle\).
cyclic \(\iff\) generator
Note: Being cyclic is a structural property.
也就是說，generator 在 isomorphic groups 之間必須一一對應。
If \(\langle a \rangle\) has a finite number of element, then the order of \(a\) is the order \(\vert \langle a \rangle \vert\) of this subgroup; otherwise, we say \(a\) is of infinite order.
經過多少次可以回到自己。
Let \(G = \langle a \rangle\) be a cyclic group. If \(a^r = a^s\) for some \(r > s\). Then \(G\) contains at most \(r - s\) elements.
Proof
Let \(m = r - s\). By Cancellation Law, we have \(a^m = e\). For any \(n \in \mathbb{Z}\), by division algorithm, we can write \(n=qm + r\) with \(0 \leq r < m\). Then \(a^n = a^{qm+r} = (a^m)^qa^r = a^r\). Therefore, every element of \(G\) is equal to one of \(a^0, \cdots, a^{m-1}\). ◼
Let \(G = \langle a \rangle\) be an infinite cyclic group. Then \(a^r \not = a^s\) for all integers \(r \not = s\). Moreover, \(G \cong \mathbb{Z}\).
Proof
If \(a^r = a^s\) for some \(r > s\), applying this theorem would lead to a contradiction.
We can define \(\phi: G \to \mathbb{Z}\) by \(\phi(a^m) = m\), which is a group isomorphism. ◼
Let \(G = \langle a \rangle\) be a cyclic group of order \(n\). Then \(G = \{a^0, a^1, \cdots, a^{n-1}\}\) and \(a^0 = a^n = e\). Moreover, \(G \cong \mathbb{Z}_n.\)
Proof
If \(a^r = a^s\) for some \(n-1 \ge r > s \ge 0\), applying this theorem would lead to a contradiction.
Now we have \(a^n = a^r\) for some \(0 \leq r \leq n-1\). In this case, \(G\) contains at most \(n - r\) elements, so we must have \(r = 0\).
Next we define \(\phi: G \to \mathbb{Z}\) by \(\phi(a^k) = \bar k\).
剩下請自行思考（homomorphism）。
Let \(a\) be an element of a group \(G\). Suppose \(a\) is of finite order \(n\). Then \(a^n = e\) and \(a^r = a^s \iff r \equiv s \mod n\).
Determine whether \(\mathbb{Z}^{\times}_{18}\) is cyclic.
Solution
We have \(\mathbb{Z}^{\times}_{18} = \{\bar 1, \bar 5, \bar 7, \overline {-7}, \overline{-5}, \overline{-1}\}\); the order of this group is \(6\). Suppose \(G = \langle a \rangle = \{e, g, g^2, g^3, g^4, g^5\}\), and the order of the elements are \(1, 6, 3, 2, 3, 6\), respectively. We now show the order of elements in \(\mathbb{Z}^{\times}_{18}\):
\[\begin{align*} \bar 1&: \text{order 1} \\ \bar 5 \to \bar 7 \to \overline{-1} \to \overline{-5} \to \overline{-7} \to \bar 1&: \text{order 6}\\ \bar 7 \to \overline{-5} \to \bar 1&: \text{order 3} \\ \overline{-7} \to \overline{-5} \to \overline{-1} \to \bar 7 \to \bar 5 \to \bar 1&: \text{order 6}\\ \overline{-5} \to \bar 7 \to \bar 1&: \text{order 3}\\ \overline{-1} \to \bar 1&: \text{order 2}.\\ \end{align*}\]We can then correspond each element of \(G\) to that of \(\mathbb{Z}^{\times}_{18}\):
\[\{e, g, g^2, g^3, g^4, g^5\} \to \{\bar 1, \bar 5, \bar 7 ,\overline{-1}, \overline{-5}, \overline{-7}\}.\]Hence we have shown that \(\mathbb{Z}^{\times}_{18}\) is cyclic. ◼
事實上，\(\vert \mathbb{Z}_{18}^{\times}\vert = 6\)，而且 \(\mathbb{Z}_{18}^{\times}\) 有 order 為 \(6\) 的 element，因此 \(\mathbb{Z}_{18}^{\times} \cong \mathbb{Z}_6\)；已知 \(\mathbb{Z}_6\) cyclic，於是所求也是 cyclic。
用這方法，只需找到一個 order 為 6 的 element，不需找出全部的 order。
Think: Suppose \(G\) is of order \(n\). If for all \(g \in G\), \(g^n = e\). Can we say that \(G\) is cyclic?
Actually this is false. Suppose \(n = 4\), we have \(G = \{e, a, b, c\}\) such that \(a^4 = b^4 = c^4 = e\). If we let \(a^2 = b^2 = c^2 = e\), we can discover that there is no single generator in \(G\). Thus if for all \(g \in G\), \(g^n = e\), \(G\) is not necessarily cyclic. (If \(G\) is indeed cyclic, this always holds, though.)
A subgroup of a cyclic group is cyclic.
Let \(G = \langle a \rangle\) be a cyclic group, and \(H\) be a subgroup. If \(H = \{e\}\), then \(H = \langle e \rangle\), which is cyclic.
Now suppose \(H \not = \{e\}\), then \(a^n \in H\) for some positive integer \(n\). Let \(m\) be the smallest positive integer such that \(a^m \in H\), and set \(c = a^m\).
We claim that \(H = \langle c \rangle\).
Since \(\langle c \rangle\) is the smallest subgroup containing \(c\), so we have \(\langle c \rangle \subseteq H\).
Let \(b \in H\). Since \(H \subset G = \langle a \rangle, b = a^n\) for some \(n \in \mathbb{Z}\). By the division algorithm, there exist integers \(q\) and \(r\) with \(0 \leq r < m\) sucht that \(n = qm + r\).
Now \(a^r = a^{n - qm} = bc^{-q} \in H\) because \(b, c \in H\).
By the assumption that \(m\) is the smallest positive integer such that \(a^m \in H\) but \(r < m\), we must have \(r = 0\). That is , \(b = c^q \in \langle c \rangle\); \(H \subseteq \langle c \rangle\).
Hence we have shown that
\[H = \langle c \rangle. \tag*{$\blacksquare$}\]具體來想，\(G = \mathbb{Z} = \langle 1 \rangle, H = \langle 5 \rangle\)，在加法之下。
The subgroups of \(\mathbb{Z}\) under addition are precisely the groups \(n\mathbb{Z}\) under addition for \(n \in \mathbb{Z}\).
\(n\mathbb{Z} = \langle n \rangle\), under addition.
Let \(G\) be a cyclic group of order \(n\) generated by \(a\). Let \(b = a^s\) and \(d = \gcd(n, s)\). Then
\[\langle b \rangle = \langle a^d \rangle = \{a^0=e, a^d, a^{2d}, \cdots, a^{n-d}\}.\]Proof
It is sufficient to show that \(a^d \in \langle b \rangle\) and \(b \in \langle a^d \rangle\).
For \(d = \gcd(n, s)\), there exist integers \(x, y\) such that \(nx + sy = d\). Thus, we have \(a^d = a^{nx+sy} = (a^n)^xa^{sy} = (e)^xa^{sy} = b^y \in \langle b \rangle\). On the other hand, we cand wirte \(s = ds'\) for some integer \(s'\). Then \(b = a^s = (a^d)^{s'} \in \langle a^d \rangle\). ◼
以下的 \(G\) 都是 cyclic group of order \(n\) generated by \(a\)。
\(\langle a^r \rangle = \langle a^s \rangle\) if and only if \(\gcd(n, r) = \gcd(n, s)\).
The other generators of \(G\) are the elements of the form \(a^r\), where \(\gcd(n ,r) = 1\).
Every subgroup of \(G\) is equal to \(\langle a^d \rangle\) for \(d\vert n\). Especially, the number of subgroups of \(G\) is equal to the number of divisors of \(n\).
\(\mathbb{Z}_{12}\) 有 \(6\) 個 subgroups。
When \(G = \mathbb{Z}_n\), every subgroup is equal to \(\langle \bar d \rangle\) for some \(d\vert n\). Moreover, the order of the subgroup is equal to \(n/d\).
\(n/d: \{0, d, 2d, \cdots, (n/d-1)d\}\).