846: For Group and Normal Subgroup, if Normal Subgroup and Quotient of Group by Normal Subgroup Are p-Groups, Group Is p-Group

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description/proof of that for group and normal subgroup, if normal subgroup and quotient of group by normal subgroup are p-groups, group is p-group

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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 quotient group of group by normal subgroup.
• The reader knows a definition of p-group.
• The reader knows a definition of subgroup generated by subset of group.
• The reader admits the proposition that for any group and its any finite-order element, the order power of the element is $1$ and the subgroup generated by the element consists of the element to the non-negative powers smaller than the element order.
• The reader admits the proposition that for any group and any element, if there is a positive natural number to power of which the element is 1 and there is no smaller such, the subgroup generated by the element consists of the element to the non-negative powers smaller than the number.

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Target Context

• The reader will have a description and a proof of the proposition that for any group and any normal subgroup, if the normal subgroup and the quotient of the group by the normal subgroup are p-groups, the group is a p-group.

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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:
$G^{\prime }$: $\in \{\text{ the groups }\}$
$G$: $\in \{\text{ the normal subgroups of }{G}^{\prime }\}$
$G^{\prime }/G$: $=\text{ the quotient group }$
$p$: $\in \{\text{ the prime numbers }\}$
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Statements:
$G,G^{\prime }/G\in \{\text{ the p-groups }\}$
$⟹$
$G^{\prime }\in \{\text{ the p-groups }\}$
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2: Proof

Whole Strategy: Step 1: take each $g^{\prime }\in G^{\prime }$ and $(G{g}^{\prime })$ and suppose that the order of $(G{g}^{\prime })$ is $p^n$, and see that $g^{\prime p^n}$ is the 1st such that $g^{\prime k}\in G$ where $k\in \mathbb{N}\setminus \{0\}$; Step 2: suppose that the order of $(g^{\prime p^n})$ is $p^m$, and see that $(g^{\prime p^n}{)}^{p^m}$ is the 1st such that $g^k=1$ where $k\in \mathbb{N}\setminus \{0\}$.

Step 1:

Let us take each $g^{\prime }\in G^{\prime }$.

Let us take $(G{g}^{\prime })\in \{\text{ the subgroups of }{G}^{\prime }/G\}$.

Let us suppose that the order of $(G{g}^{\prime })$ is $p^n$: $G^{\prime }/G$ is a p-group.

$(G{g}^{\prime })=\{1,G{g}^{\prime },...,G{g}^{\prime p^n-1}\}$ where $G{g}^{\prime p^n}=1$, by the proposition that for any group and its any finite-order element, the order power of the element is $1$ and the subgroup generated by the element consists of the element to the non-negative powers smaller than the element order.

That means that $g^{\prime p^n}$ is the 1st such that $g^{\prime k}\in G$ where $k\in \mathbb{N}\setminus \{0\}$: $G{g}^{\prime k}=1$ means that $g^{\prime k}\in G$.

Step 2:

Let us think of $(g^{\prime })$.

$(g^{\prime })=\{1,g^{\prime },g^{\prime -1},g^{\prime 2},g^{\prime -2},...\}$, which may have some duplications.

We already know that $g^{\prime p^n}\in G$.

Let us suppose that the order of $(g^{\prime p^n})$ is $p^m$: $G$ is a p-group.

$(g^{\prime p^n})=\{1,g^{\prime p^n},...,(g^{\prime p^n}{)}^{p^m-1}\}$ where $(g^{\prime p^n}{)}^{p^m}=1$, by the proposition that for any group and its any finite-order element, the order power of the element is $1$ and the subgroup generated by the element consists of the element to the non-negative powers smaller than the element order.

That means that $(g^{\prime p^n}{)}^{p^m}$ is the 1st such that $(g^{\prime p^n}{)}^k=1$ where $k\in \mathbb{N}\setminus \{0\}$.

Let us suppose that $g^{\prime k}=1$ where $k\in \mathbb{N}\setminus \{0\}$.

Let $k=l{p}^n+j$ where $l\in \mathbb{N}$ and $j\in \mathbb{N}$ such that $0\le j<p^n$.

$g^{\prime k}=g^{\prime l{p}^n+j}=g^{\prime l{p}^n}{g}^{\prime j}=1$, which implies that $j=0$, because $g^{\prime j}=(g^{\prime l{p}^n}{)}^{-1}=g^{\prime -l{p}^n}=(g^{\prime p^n}{)}^{-l}\in G$, but $g^{\prime p^n}$ is the 1st such that $g^{\prime k}\in G$ where $k\in \mathbb{N}\setminus \{0\}$.

So, $k=l{p}^n$ where $l\in \mathbb{N}\setminus \{0\}$.

But $p^m\le l$, because $g^k=g^{l{p}^n}=(g^{p^n}{)}^l$, but $(g^{\prime p^n}{)}^{p^m}$ is the 1st such that $(g^{\prime p^n}{)}^k=1$ where $k\in \mathbb{N}\setminus \{0\}$.

$(g^{\prime p^n}{)}^{p^m}=g^{\prime p^n{p}^m}=g^{\prime p^{n+m}}=1$.

So, $(g^{\prime })=\{1,g^{\prime },...,g^{\prime p^{n+m}-1}\}$, by the proposition that for any group and any element, if there is a positive natural number to power of which the element is 1 and there is no smaller such, the subgroup generated by the element consists of the element to the non-negative powers smaller than the number.

So, the order of $g^{\prime }$ is $p^{n+m}$, which means that $G^{\prime }$ is a p-group.

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References

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