786: Abelian Group Is Simple Group iff Its Order Is Prime Number

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description/proof of that Abelian group is simple group iff its order is prime 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: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of Abelian group.
• The reader knows a definition of simple group.
• The reader admits Lagrange's theorem.

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Target Context

• The reader will have a description and a proof of the proposition that any Abelian group is a simple group if and only if its order is a prime 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:
$G$: $\in \{\text{ the Abelian groups }\}$
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Statements:
$G\in \{\text{ the simple groups }\}$
$⟺$
$|G|=p$ where $p\in \{\text{ the prime numbers }\}$
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2: Natural Language Description

Any Abelian group is a simple group if and only if the order, $|G|$ is any prime number, $p$.

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3: Proof

Whole Strategy: Step 1: suppose that $G$ is a simple group, take any element, $g\in G\setminus \{1\}$, and the subgroup generated by $g$, $(g)$, and see that $(g)=G$; Step 2: when $g^2=1$, see that $|G|=2$, otherwise, take $(g^2)$ and see that $(g^2)=G=(g)$; Step 3: see that Step 1 and Step 2 imply that $|G|$ is finite; Step 4: let $|G|=pk$ for a prime number, $p$, and see that $k=1$; Step 5: suppose that $|G|=p$ for a prime number, $p$, and see that its subgroups are only $\{1\}$ and $G$.

Step 1:

Let us suppose that $G$ is a simple group.

Let us take any element, $g\in G\setminus \{1\}$, and the subgroup generated by $g$, $(g)=\{g^{\prime }\in G|g^{\prime }=g^z,z\in \mathbb{Z}\}$.

$(g)$ is indeed a subgroup: $g^z{g}^{z^{\prime }}=g^{z+z^{\prime }}\in (g)$; $g^0=1\in (g)$; for each $g^z\in (g)$, $g^z{g}^{-z}=g^{-z}{g}^z=g^0=1$.

$(g)$ is a normal subgroup, because each subgroup of any Abelian group is a normal subgroup.

So, $(g)=G$, because $G$ is simple (obviously, $G\ne \{1\}$).

Step 2:

When $g^2=1$, $G=\{1,g\}$, and $|G|=2$, which is a prime number.

Let us suppose otherwise hereafter.

Let us take $(g^2)$, which is a normal subgroup of $G$, as before, because we just have took $g^2$ instead of $g$.

$(g^2)=G$, as before.

Step 3:

Let us see that $G=(g)=(g^2)$ implies that $|G|$ is finite.

There is a $z^{\prime }$ such that $g=g^{2z^{\prime }}$.

$1=g{g}^{-1}=g^{2z^{\prime }}{g}^{-1}=g^{2z^{\prime }-1}$. $2z^{\prime }-1\ne 0$, because it is odd. When $0<2z^{\prime }-1$, let $n:=2z^{\prime }-1$; otherwise, $0<-(2z^{\prime }-1)$, and let $n:=-(2z^{\prime }-1)$. Anyway, $g^n=1$, because $1=1^{-1}=(g^{2z^{\prime }-1}{)}^{-1}=g^{-(2z^{\prime }-1)}$. $n\ne 1$, because $g^1=g\ne 1$. $g^{-1}=g^{n-1}$, because $g{g}^{n-1}=g^{n-1}g=g^n=1$.

So, each element of $(g)$ equals one of $\{1,g,...,g^{n-1}\}$ (there may be some duplications in the set), because for each $z\in \mathbb{Z}$, $z=nk+l$ where $0\le l<n$ and $g^z=g^{nk+l}=g^{nk}{g}^l=(g^n{)}^k{g}^l=1^k{g}^l=g^l$.

So, $|G|=|(g)|$ is finite.

Step 4:

Let us take the prime factorization of $|G|$. As $1<|G|$, there is at least 1 prime number, $p$, such that $|G|=pk$ where $k\in \mathbb{N}\setminus \{0\}$.

As $G=(g)$, $G=\{1,g,...,g^{pk-1}\}$ where $g^{pk}=1$.

Let us take $(g^p)$. As $g^{pk}=1$, $(g^p)=\{1,g^p,...,g^{p(k-1)}\}$.

As $(g^p)$ is a normal subgroup as before and $G$ is simple, $(g^p)$ is $\{1\}$ or $G$. But it cannot be $G$, because at least $g,...,g^{p-1}$ are missing there. So, $(g^p)=\{1\}$. That means that $k=1$.

So, $|G|=p$.

Step 5:

Let us suppose that $|G|=p$ for a prime number, $p$.

By Lagrange's theorem, each of the only possible subgroups of $G$ has order $1$ or $p$. So, the subgroups of $G$ are only $\{1\}$ and $G$. The normal subgroups of $G$ are $\{1\}$ and $G$, even more.

So, $G$ is simple.

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References

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