987: For Finite Group, if There Is at Most 1 Subgroup for Each Divisor of Order of Group, Group Is Cyclic

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description/proof of that for finite group, if there is at most 1 subgroup for each divisor of order of group, group is cyclic

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Topics

About: group

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The table of contents of this article

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

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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.

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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.

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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.

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Main Body

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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:
$\mathrm{\forall }d|n(|\{\text{ the }d\text{ -ordered subgroups of }G\}|\le 1)$
$⟹$
$G\in \{\text{ the cyclic groups }\}$
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2: Proof

Whole Strategy: Step 1: divide $G$ by the orders of the elements as $\{S_d|d|n\}$, and see that $n=|G|=\sum _{d|n}|S_d|$; Step 2: see that $|S_d|\le \varphi (d)$; Step 3: conclude the proposition.

Step 1:

Let us divide $G$ by the orders of the elements as $\{S_d|d|n\}$, 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, $|G|=\sum _{d|n}|S_d|$.

Step 2:

Let us see that $|S_d|\le \varphi (d)$.

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 $\varphi :\mathbb{N}\setminus \{0\}\to \mathbb{N}\setminus \{0\}$, $\varphi (d)$ is the number of the single-element-generators of $G_d$, by Note for the definition of Euler's totient function, and so, $|S_d|\le \varphi (d)$, 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:

$n=|G|=\sum _{d|n}|S_d|\le \sum _{d|n}\varphi (d)=n$, 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 $d|n$, $|S_d|=\varphi (d)$ after all, because if $|S_d|<\varphi (d)$ for a $d$, $n=\sum _{d|n}|S_d|<\sum _{d|n}\varphi (d)=n$, a contradiction.

Especially, $|S_n|=\varphi (n)$, but $1\le \varphi (n)$, because $gcd(n,1)=1$.

That means that $G$ is the cyclic group generated by an element of $S_n$.

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References

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