2.8 Cyclic Groups and Orders of Elements
For this section \(G\) is a group with identity element \(e\).
Suppose \(g\in G\). Set \(K(g) = \{ n\in \mathbb {Z}: g^n = e\} \) and set \(K^+(g) = K(g)\cap \mathbb {P}\). We say that \(g\) has infinite order and write \(|g| = \infty \) if \(K^+(g) = \emptyset \). Otherwise, the order of \(g\) is the smallest element of \(K^+(g)\).
For each \(g\in G\), \(K(g)\leq \mathbb {Z}\). Consequently, \(K(g) = d\mathbb {Z}\) for some (uniquely defined) nonnegative integer \(d\).
If \(K(g) = \{ 0\} \), then \(d = 0\) and \(|g| = \infty \). Otherwise, \(d\) is the smallest element of \(K^+(g) = d\mathbb {Z}\cap \mathbb {P}\). So \(|g| = d\).
We use the one-step subgroup test. For this, first note that \(0\in K(g)\) as \(g^0 = e\) by definition. On the other hand, suppose \(n,m\in K(g)\). Then \(g^{n-m} = g^n (g^m)^{-1} = e e^{-1} = e\). So \(n-m\in K(g)\) as well.
This shows that \(K(g) \leq \mathbb {Z}\). The rest follows from Corollary 2.47 and Definintion 2.50.
Suppose \(g\in G\), and suppose \(i\) and \(j\) are integers.
If \(|g| = \infty \), then \(g^i = g^j \Leftrightarrow i = j\).
If \(|g| = d {\lt} \infty \), then \(g^i = g^j \Leftrightarrow d | i-j\).
Without loss of generality, we can suppose that \(i \leq j\).
First assume that \(|g| = \infty \). Then, \(K(g) = \{ 0\} \) by Proposition 2.51. So \(g^i = g^j \Rightarrow g^{i-j} = g^i (g^j)^{-1} = e \Rightarrow i-j = 0\Rightarrow i = j\).
On the other hand, suppose \(|g| = d\) for some nonzero integer \(d\). Then, by Proposition 2.51, \(K(g) = d\mathbb {Z}\). So \(g^i = g^j \Rightarrow g^{i-j} = g^i (g^j)^{-1} = e \Rightarrow i-j \in d\mathbb {Z}\Rightarrow i-j = dk\) for some \(k\in \mathbb {Z}\).
With the notation as in 2.52, suppose \(d = |g| {\gt} \infty \) and \(i, j\in \mathbb {Z}\). Then \(g^i = g^j\Leftrightarrow i \equiv j \pmod{d}\).
Obvious.
The order \(|G|\) of \(G\) is \(\infty \) if \(G\) is infinite and the number of elements of \(G\) otherwise.
Suppose \(G = \langle g\rangle \) is cyclic.
If \(|g| = \infty \), then \(|G| = \infty \) and all the elements \(g^i\) with \(i\in \mathbb {Z}\) are distinct.
If \(|g| = d {\lt}\infty \), then \(|G| = d\) and \(G = \{ e, g, \ldots , g^{d-1}\} \). In other words, every element of \(G\) has the form \(g^r\) for some unique integer \(r\) satisfying \(0\leq r {\lt} d\).
1 follows directly from Theorem 2.52.
To see 2, suppose \(h\in G\). Since \(G = \langle g\rangle \), we have \(h = g^i\) for some \(i\in \mathbf{Z}\). Then, using the division algorithm, we can write \(i = qd + r\), where \(q,r\in \mathbb {Z}\) and \(0\leq r {\lt} d\). Then \(i\equiv r \pmod{d}\). So \(h = g^r\) as well. Thus, \(h\in \{ e, g,\ldots , g^{d-1}\} \).
On the other hand, if \(g^i = g^j\) with \(0\leq i\leq j {\lt} d\), then \(d | j-i\) by Theorem 2.52. So \(i = j\).
Suppose \(n\) and \(m\) are integers which are not both \(0\), and set \(d = \gcd (n,m)\). Then \(\gcd (n/d, m/d) = 1\).
Since \(n\) and \(m\) are not both \(0\), \(d \neq 0\). So suppose \(e | (n/d)\) and \(e | (m/d)\). Then \(ed|n\) and \(ed| m\). So, by the definition of the gcd, \(ed | d\). but this iplies that \(e|1\). So \(e = \pm 1\). And this shows that \(\gcd (n/d, m/d) = 1\).
Suppose \(G = \langle g \rangle \) is a cyclic group and \(i\in \mathbb {Z}\).
If \(|g| = \infty \), then
\[ |g^i| = \begin{cases} \infty , & i \neq 0;\\ 1, & i = 0. \end{cases} \]If \(|g| = n {\lt} \infty \), then
\[ |g^i| = \frac{n}{\gcd (n,i)}. \]
1 Exercise.
2 Suppose \(|g| = n {\lt} \infty \). Then, by Proposition 2.51, \(K(g) = n\mathbb {Z}\). So \(g^k = e\Leftrightarrow n|k\). Consequently, \((g^{i})^k = g^{ik} = e \Leftrightarrow n | ik\). Seting \(d = \gcd (n,i)\), we see that \(n | ik \Leftrightarrow (n/d) | (i/d) k\). But \(\gcd (n/d, i/d) = 1\). So this happens if and only if \((n/d) | k\). Consequently, \(K(g^i) = (n/d)\mathbb {Z}\). So \(|g^i| = n/d\).