description/proof of that for finite group, if there is at most 1 subgroup for each divisor of order of group, group is cyclic
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of cyclic group by element.
- The reader knows a definition of Euler's totient function.
- The reader admits Lagrange's theorem.
- The reader admits the proposition that any positive natural number is the sum of the Euler's totient function results of the divisors of the number.
Target Context
- The reader will have a description and a proof of the proposition that for any finite group, if there is at most 1 subgroup for each divisor of the order of the group, the group is cyclic.
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:
\(G\): \(\in \{\text{ the } n \text{ -ordered finite groups }\}\)
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Statements:
\(\forall d \vert n (\vert \{\text{ the } d \text{ -ordered subgroups of } G\} \vert \le 1)\)
\(\implies\)
\(G \in \{\text{ the cyclic groups }\}\)
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2: Proof
Whole Strategy: Step 1: divide \(G\) by the orders of the elements as \(\{S_d \vert d \vert n\}\), and see that \(n = \vert G \vert = \sum_{d \vert n} \vert S_d \vert\); Step 2: see that \(\vert S_d \vert \le \phi (d)\); Step 3: conclude the proposition.
Step 1:
Let us divide \(G\) by the orders of the elements as \(\{S_d \vert d \vert n\}\), where \(S_d\) is the subset of \(G\) whose elements have the order \(d\).
As each element of \(G\) has the definite order and the order is a divisor of \(n\) by Lagrange's theorem, that is indeed a disjoint division of \(G\).
So, \(\vert G \vert = \sum_{d \vert n} \vert S_d \vert\).
Step 2:
Let us see that \(\vert S_d \vert \le \phi (d)\).
Each element of \(S_d\) generates a \(d\)-ordered subgroup of \(G\). But those subgroups are the same subgroup, \(G_d\), by the supposition of this proposition.
\(G_d\) is a cyclic group and each element of \(S_d\) is in \(G_d\) and generates \(G_d\).
For Euler's totient function \(\phi: \mathbb{N} \setminus \{0\} \to \mathbb{N} \setminus \{0\}\), \(\phi (d)\) is the number of the single-element-generators of \(G_d\), by Note for the definition of Euler's totient function, and so, \(\vert S_d \vert \le \phi (d)\), because each element of \(S_d\) is a generator of \(G_d\) although not all the single-element-generators may be in \(S_d\).
Step 3:
\(n = \vert G \vert = \sum_{d \vert n} \vert S_d \vert \le \sum_{d \vert n} \phi (d) = n\), by the proposition that any positive natural number is the sum of the Euler's totient function results of the divisors of the number.
Then, for each \(d \vert n\), \(\vert S_d \vert = \phi (d)\) after all, because if \(\vert S_d \vert \lt \phi (d)\) for a \(d\), \(n = \sum_{d \vert n} \vert S_d \vert \lt \sum_{d \vert n} \phi (d) = n\), a contradiction.
Especially, \(\vert S_n \vert = \phi (n)\), but \(1 \le \phi (n)\), because \(gcd (n, 1) = 1\).
That means that \(G\) is the cyclic group generated by an element of \(S_n\).
