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
- 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:
\(G \in \{\text{ the cyclic groups }\}\)
\(\land\)
\(\forall g \in G \setminus \{1\} (\langle g \rangle = G)\)
//
2: Proof
Whole Strategy: Step 1: take each \(g \in G \setminus \{1\}\) and see that \(\langle g \rangle = G\), by Lagrange's theorem.
Step 1:
Let \(g \in G \setminus \{1\}\) be any: there is such a \(g\), because \(2 \le \vert G \vert = p\).
\(\langle g \rangle\) is a subgroup of \(G\), and by Lagrange's theorem, \(\vert \langle g \rangle \vert = p \text{ or } 1\). But it cannot be \(1\), because that would mean \(g = 1\). So, \(\vert \langle g \rangle \vert = p\).
So, \(\langle g \rangle = G\), which means that \(G\) is cyclic.
As \(g\) is arbitrary, each element of \(G\) except \(1\) generates \(G\).
