989: Prime-Number-Ordered Group Is Cyclic and Each Element Except 1 Generates Group

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

description/proof of that prime-number-ordered group is cyclic and each element except 1 generates group

---

Topics

About: group

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

---

Starting Context

• The reader knows a definition of cyclic group by element.
• The reader admits Lagrange's theorem.

---

Target Context

• The reader will have a description and a proof of the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group.

---

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:
$p$: $\in \{\text{ the prime numbers }\}$
$G$: $\in \{\text{ the }p\text{ -ordered groups }\}$
//

Statements:
$G\in \{\text{ the cyclic groups }\}$
$\wedge$
$\mathrm{\forall }g\in G\setminus \{1\}(⟨g⟩=G)$
//

---

2: Proof

Whole Strategy: Step 1: take each $g\in G\setminus \{1\}$ and see that $⟨g⟩=G$, by Lagrange's theorem.

Step 1:

Let $g\in G\setminus \{1\}$ be any: there is such a $g$, because $2\le |G|=p$.

$⟨g⟩$ is a subgroup of $G$, and by Lagrange's theorem, $|⟨g⟩|=p\text{ or }1$. But it cannot be $1$, because that would mean $g=1$. So, $|⟨g⟩|=p$.

So, $⟨g⟩=G$, which means that $G$ is cyclic.

As $g$ is arbitrary, each element of $G$ except $1$ generates $G$.

---

References

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