description/proof of that for finite group, if for each divisor of order of group, there are at most divisor elements of orders that divide divisor, 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 for each divisor of the order of the group, there are at most the divisor elements of orders that divide the divisor, 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:
\(\forall d \vert n (\vert \{g \in G \vert \text{ } \vert \langle g \rangle \vert \text{ } \vert \text{ } d\} \vert \le d)\)
\(\implies\)
\(G \in \{\text{ the cyclic groups }\}\)
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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 \(S_d\), \(g_j\), generates the \(d\)-ordered subgroup of \(G\), \(\{g_j^1, ..., g_j^d = 1\}\).
For each \(1 \le k \le d\), \((g_j^k)^d = g_j^{k d} = (g_j^d)^k = 1^k = 1\), which means that \(g_j^k\) has an order that divides \(d\).
So, the at most \(d\) elements whose orders divide \(d\) the supposition of this proposition mentions are nothing but all the elements of \(\{g_j^1, ..., g_j^d = 1\}\).
As there are \(\{g_1^1, ..., g_1^d = 1\}\), \(\{g_2^1, ..., g_2^d = 1\}\), ..., in fact, those subgroups are the same, because otherwise, there would be more than \(d\) elements whose orders divided \(d\).
As each \(g_j\) generates \(\{g_1^1, ..., g_1^d = 1\}\), each \(g_j\) is in \(\{g_1^1, ..., g_1^d = 1\}\), and the elements of \(S_d\) are some generators of \(\{g_1^1, ..., g_1^d = 1\}\).
\(\{g_1^1, ..., g_1^d = 1\}\) is cyclic, and so, for Euler's totient function, \(\phi: \mathbb{N} \setminus \{0\} \to \mathbb{N} \setminus \{0\}\), \(\vert S_d \vert \le \phi (d)\), because \(\phi (d)\) is the number of the single-element-generators of \(\{g_1^1, ..., g_1^d = 1\}\), by Note for the definition of Euler's totient function.
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 \(S_n\).
