description/proof of that for finite cyclic group and its prime factor of order, cyclic subgroup of factor order can be extracted in certain way
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of cyclic group by element.
Target Context
- The reader will have a description and a proof of the proposition that for any finite cyclic group and any prime factor of the order of the group, a cyclic subgroup of the factor order can be extracted in a certain way.
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:
\(p\): \(\in \{\text{ the prime numbers }\}\)
\(G\): \(\in \{\text{ the cyclic groups }\}\) with order \(p n\), \(= \{g , ..., g^{p n} = 1\}\)
//
Statements:
\(G_p := \{g^n, g^{2 n}, ..., g^{p n}\} \in \{\text{ the } p \text{ -ordered cyclic groups } \}\)
//
2: Note
By Cauchy's theorem, a \(p\)-ordered subgroup (inevitably cyclic, by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group) can be extracted without \(G\) required to be cyclic; the point of this proposition is that a \(p\)-ordered subgroup can be extracted in the specific way, not "a \(p\)-ordered subgroup is hidden somewhere unknown".
This proposition does not claim that \(G_p\) is the only \(p\)-ordered subgroup, but it is indeed the only \(p\)-ordered subgroup, by the proposition that for any cyclic group and its any prime factor of the order of the group, there is at most 1 subgroup of the factor order.
3: Proof
Whole Strategy: Step 1: see that \(G_p\) is indeed a \(p\)-ordered subgroup.
Step 1:
Let us see that \(G_p\) is a subgroup.
For any \(g^{j n}, g^{k n} \in G_p\), \(g^{j n} g^{k n} = g^{(j + k) n}\). \(j + k = m p + l\) for an \(m, l \in \mathbb{N}\) such that \(0 \le l \lt p\). \(g^{(j + k) n} = g^{(m p + l) n} = g^{m p n + l n} = g^{m p n} g^{l n} = (g^{p n})^m g^{l n} = 1^m g^{l n} = 1 g^{l n} = g^{l n}\). When \(0 \lt l \lt p\), \(g^{l n} \in G_p\). When \(l = 0\), \(g^{l n} = g^0 = 1 = g^{p n} \in G_p\). So, \(g^{j n} g^{k n} \in G_p\) anyway.
\(g^{p n} = 1 \in G_p\).
For each \(g^{j n} \in G_p\), \(g^{(p - j) n} \in G_p\), and \(g^{j n} g^{(p - j) n} = g^{j n + (p - j) n} = g^{p n} = 1\); \(g^{(p - j) n} g^{j n} = g^{j n + (p - j) n} = g^{p n} = 1\).
\(G_p\) is \(p\)-ordered, because \(\{g^n, g^{2 n}, ..., g^{p n}\}\) is distinct, because \(\{g , ..., g^{p n} = 1\}\) is distinct.
