definition of cyclic group by element
Topics
About: group
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Note
Starting Context
- The reader knows a definition of group.
Target Context
- The reader will have a definition of cyclic group by element.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( p\): \(\in \{\text{ the objects }\}\)
\(*\langle p \rangle\): \(= \{p^j \vert j \in \mathbb{Z}\}\) or \(= \{p^j \vert j \in \mathbb{N}, 0 \le j \le (n - 1)\}\) for an \(n \in \mathbb{N} \setminus \{0\}\), with \(p^j p^k = p^{j + k}\) or \(p^j p^k = p^{[j + k]}\) as the group operation, where \([j + k]\) denote the residue of \(j + k\) divided by \(n\)
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Conditions:
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What does "object" mean? Well, 'object' is any object in the universe. Someone who bases the whole mathematics on the ZFC set theory will say that it is a set, but we are not such a someone.
Someone may wonder what \(p^j\) means where \(p\) is just an object. Well, it is just a symbol we have invented now: of course, \(p^j\) could not be define like \(p ... p\), because the group operation has not been defined yet: the group operation is defined after the set of the group has been defined.
Then, after the group operation has been defined, \(p^j = p ... p\) as a result.
2: Natural Language Description
For any object, \(p\), \(\langle p \rangle\): \(= \{p^j \vert j \in \mathbb{Z}\}\) or \(= \{p^j \vert j \in \mathbb{N}, 0 \le j \le (n - 1)\}\) for an \(n \in \mathbb{N} \setminus \{0\}\), with \(p^j p^k = p^{j + k}\) or \(p^j p^k = p^{[j + k]}\) as the group operation, where \([j + k]\) denote the residue of \(j + k\) divided by \(n\)
3: Note
\(\langle p \rangle\) is indeed a group for the infinite group case: 0) \(p^j p^k = p^{j + k} \in \langle p \rangle\); 1) \((p^j p^k) p^l = p^{j + k} p^l = p^{j + k + l} = p^j p^{k + l} = p^j (p^k p^l)\); 2) \(p^0\) is the identity element, because \(p^0 p^j = p^{0 + j} = p^j = p^{j + 0} = p^j p^0\), so, called \(1\); 3) for each \(p^j\), \(p^{-j}\) is the inverse, because \(p^j p^{-j} = p^{j + - j} = p^0 = 1 = p^0 = p^{-j + j} = p^{-j} p^j\).
\(\langle p \rangle\) is indeed a group for the finite group case: 0) \(p^j p^k = p^{[j + k]} \in \langle p \rangle\); 1) \((p^j p^k) p^l = p^{[j + k]} p^l = p^{[j + k + l]} = p^j p^{[k + l]} = p^j (p^k p^l)\); 2) \(p^0\) is the identity element, because \(p^0 p^j = p^{[0 + j]} = p^j = p^{[j + 0]} = p^j p^0\), so, called \(1\); 3) for each \(p^j\), \(p^{n - j}\) is the inverse, because \(p^j p^{n - j} = p^{[j + n - j]} = p^0 = 1 = p^0 = p^{[n - j + j]} = p^{n - j} p^j\).
Do not make the confusion for the infinite case like that only \(p, p^2, ...\) constitutes the group: the sequence never returns to \(1\) or produces the inverse of \(p\).
In many cases, for a group, \(G\), the cyclic subgroup of \(G\) by an element, \(p \in G\), is taken. That is indeed a cyclic group by \(p\) by this definition by identifying the symbol \(p^j\) with the multiplication, \(p ... p\), on \(G\). Whether it is an infinite group or an \(n\)-order group depends on \(p\). When \(G\) is finite, it inevitably becomes finite, but when \(G\) is infinite, it may be or may not be infinite.
