712: Cyclic Group by Element

<The previous article in this series | The table of contents of this series | The next article in this series>

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:
$p$: $\in \{\text{ the objects }\}$
$\ast ⟨p⟩$: $=\{p^j|j\in \mathbb{Z}\}$ or $=\{p^j|j\in \mathbb{N},0\le j\le (n-1)\}$ 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 $n$
//

Conditions:
//

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, $p$, $⟨p⟩$: $=\{p^j|j\in \mathbb{Z}\}$ or $=\{p^j|j\in \mathbb{N},0\le j\le (n-1)\}$ 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

$⟨p⟩$ is indeed a group for the infinite group case: 0) $p^j{p}^k=p^{j+k}\in ⟨p⟩$; 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 $p^j{p}^{-j}=p^{j+-j}=p^0=1=p^0=p^{-j+j}=p^{-j}{p}^j$.

$⟨p⟩$ is indeed a group for the finite group case: 0) $p^j{p}^k=p^{[j+k]}\in ⟨p⟩$; 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.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>