845: For Finite p-Group, for Natural Number Smaller Than Power to Which p Is Order of Group, There Is Normal Subgroup of Group Whose Order Is p to Power of Natural Number

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description/proof of that for finite $p$-group, for natural number smaller than power to which $p$ is order of group, there is normal subgroup of group whose order is $p$ to power of natural number

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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 $p$-group.
• The reader knows a definition of normal subgroup of group.
• The reader knows a definition of quotient group of group by normal subgroup.
• The reader knows a definition of center of group.
• The reader knows a definition of union of set.
• The reader admits the proposition that for any finite group, the group is a $p$-group if and only if the order of the group is $p$ to the power of a natural number.
• The reader admits the proposition that for any $p$-group, its center is not trivial.
• The reader admits Lagrange's theorem.
• The reader admits Cauchy's theorem.
• The reader admits the correspondence theorem for group and quotient group.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite $p$-group, for each natural number smaller than the power to which $p$ is the order of the group, there is a normal subgroup of the group whose order is $p$ to the power of the natural number.

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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:
$G$: $\in \{\text{ the finite }p\text{ -groups }\}$
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Statements:
$\mathrm{\exists }n\in \mathbb{N}(|G|=p^n)$
$\wedge$
$\mathrm{\forall }k\in \mathbb{N}\text{ such that }0\le k<n(\mathrm{\exists }{G}_k\in \{\text{ the normal subgroups of }G\}(|G_k|=p^k))$
//

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2: Proof

Whole Strategy: prove it inductively; Step 1: admit that $|G|=p^n$; Step 2: prove it for $k=0$; Step 3: suppose that there are some $G_0,...,G_{l-1}$ and see that there is a $G_l$.

Step 1:

For any finite $p$-group, there is an $n\in \mathbb{N}$ such that $|G|=p^n$, by the proposition that for any finite group, the group is a $p$-group if and only if the order of the group is $p$ to the power of a natural number, which is not the main point of this proposition but is a preparation.

Step 2:

Let us suppose that $k=0$.

$G_0:=\{1\}$ is the normal subgroup of $G$ such that $|G_0|=p^0=1$.

Step 3:

Step 3 Strategy: Step 3-1: suppose that there are $G_0,...,G_{l-1}$; Step 3-2: take $G/G_{l-1}$ and see that it is a $p$-group; Step 3-3: take $Z(G/G_{l-1})$ and see that it is a $p$-group; Step 3-4: take a subgroup of $Z(G/G_{l-1})$, $N$, such that $|N|=p$ and see that $N$ is a normal subgroup of $G/G_{l-1}$; Step 3-5: see that $G_l:=\cup N$ is a normal subgroup of $G$ such that $|G_l|=p^l$.

Step 3-1:

Let us suppose that there are $G_0,...,G_{l-1}$.

Step 3-2:

Let us think of the quotient, $G/G_{l-1}$, which is valid because $G_{l-1}$ is a normal subgroup of $G$.

$|G/G_{l-1}|=p^n/p^{l-1}=p^{n-(l-1)}$, by the definition of quotient group of group by normal subgroup.

So, $G/G_{l-1}$ is a $p$-group, by the proposition that for any finite group, the group is a $p$-group if and only if the order of the group is $p$ to the power of a natural number.

Step 3-3:

Let us think of the center of $G/G_{l-1}$, $Z(G/G_{l-1})$.

$Z(G/G_{l-1})$ is a normal subgroup of $G/G_{l-1}$.

$1<|Z(G/G_{l-1})|$, by the proposition that for any $p$-group, its center is not trivial, which means that $|Z(G/G_{l-1})|=p^m$ for an $m\in \mathbb{N}\setminus \{0\}$, by Lagrange's theorem, which means that $Z(G/G_{l-1})$ is a $p$-group by the proposition that for any finite group, the group is a $p$-group if and only if the order of the group is $p$ to the power of a natural number.

Step 3-4:

There is a subgroup of $Z(G/G_{l-1})$, $N$, such that $|N|=p$, by Cauchy's theorem.

$N$ is a normal subgroup of $G/G_{l-1}$, because $N$ is a subgroup of the center: for each $g\in G/G_{l-1}$, $gN{g}^{-1}=N$, because for each $g^{\prime }\in N$, $g{g}^{\prime }{g}^{-1}=g^{\prime }$, because $g^{\prime }\in Z(G/G_{l-1})$.

Step 3-5:

Then, $\cup N$ is a normal subgroup of $G$, by the correspondence theorem for group and quotient group.

$|\cup N|=|G_{l-1}||N|=p^{l-1}p=p^l$.

So, we can take $G_l:=\cup N$.

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References

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