Cyclic Group
Table of Contents
Every cyclic group is abelian.
1. Generators of Cyclic Groups
Let \(G\) be a group and \(a \in G\)
- If \(a\) has infinite order, then all distinct power of \(a\) are distinct group elements
- If \(a\) has finite order, then \( \lang a \rang = \{ e,a,a^{2}, \dots, a^{n-1} \} \) and \( a^{i} = a^{j} \iff n \mid i-j \)
Let \(o(a) = n\), then \(a^{k}=e \implies n \mid k\)
The order of a cyclic group is equal to the order of its generator.
If \(a \in \lang a^{k} \rang\), then \(\lang a \rang \sube \lang a^{k} \rang\)
Let \(G = \lang a \rang\) be a cyclic group of order \(n\). Then, \(G = \lang a^{k} \rang\) if and only if \(\gcd(k, n) = 1\).
2. Subgroups of Cyclic Groups
Fundamental Theorem of Cyclic Groups
- Every subgroup of a cyclic group is cyclic.
- If \( o(\lang a \rang) = n \), then the order of any subgroup of \(\lang a \rang\) is a divisor of \(n\)
- For each positive divisor \(k\) of \(n\), the group \(\lang a \rang\) has exactly one subgroup of order \(k\), that is \(\lang a^{n/k} \rang\).
Let \(G\) be a finite cyclic group. Then, the number of distinct subgroups of \(G\) is equal to the number of distinct factors of \(o(G)\).
For each positive divisor \(k\) of \(n\), the set \(\lang n/k \rang\) is the subgroup of \(\mathbb{Z}_{n}\) of order \(k\).
Number of generators of a finite cyclic group of order \(n\) is \(\phi(n)\).
If \(m\) is a positive divisor of \(n\), then the number of elements of order \(m\) in a cyclic group of order \(n\) is \(\phi(m)\).
Solve Cyclic Group Related Questions
I find extended Euclidean theorem is always helpful in solving problems related to \(\gcd\) and cyclic groups.
If \(o(\lang a \rang)=n\), prove that \(\lang a^{k} \rang = \lang a^{\gcd(n,k)} \rang\)
Let \(\gcd(n,k) = d\), then \(k = df, n = dg, f \in \mathbb{Z}, g \in \mathbb{Z}\). And as \(o(\lang a \rang) = n\), so \(a^{n}=e\)
For any \(x \in \lang a^{k} \rang\), then \(x = a^{kt}, t \in \mathbb{Z}\), therefore \[ x = a^{dft} = (a^{d})^{ft} \implies x \in \lang a^{d} \rang \] Thus \(\lang a^{k} \rang \sube \lang a^{d} \rang\).
For any \(x \in \lang a^{d} \rang\), then \(x = a^{dt}, t \in \mathbb{Z}\). By extended Euclidean theorem, since \(d = \gcd(n,k)\), there exists \(y, z \in \mathbb{Z}\) such that \(d = yn+zk\). Therefore, \[ x=a^{t(yn+zk)}=(a^{n})^{ty}\cdot (a^{k})^{tz}=(a^{k})^{tz} \implies x \in \lang a^{k} \rang \] Thus \(\lang a^{d} \rang \sube \lang a^{k} \rang\).
Therefore, \(\lang a^{d} \rang = \lang a^{k} \rang\)