943: For Finite Group and Subgroup Whose Index Is Smallest Prime Divisor of Order of Group, Subgroup Is Normal Subgroup

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description/proof of that for finite group and subgroup whose index is smallest prime divisor of order of group, subgroup is normal subgroup

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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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of normal subgroup of group.
• The reader admits the proposition that for any group, any subgroup, and the set of the conjugates of the subgroup, there is a group homomorphism from the group into the permutations group of the conjugates set, whose (the homomorphism's) kernel is a subgroup of the normalizer of the subgroup on the group.
• The reader admits Lagrange's theorem.
• The reader admits the proposition that for any group homomorphism, the quotient group of the domain by the kernel of the homomorphism is 'groups - homomorphisms' isomorphic to the range of the homomorphism: the 1st isomorphism theorem for groups.
• The reader admits the proposition that for any group, any subgroup, and the left or right coset of the subgroup by any element of the group, the conjugates of the subgroup by any elements or the inverses of any elements respectively of the coset are the same.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite group and any subgroup whose index is the smallest prime divisor of the order of the group, the subgroup is a normal subgroup.

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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:
$G^{\prime }$: $\in \{\text{ the groups }\}$
$|G^{\prime }|$: $=p_1^{n_1}...p_k^{n_k}$, where $p^j\in \{\text{ the prime numbers }\}$ and $n_j\in \mathbb{N}\setminus \{0\}$ such that $p_1<...<p_k$
$G$: $\in \{\text{ the subgroups of }{G}^{\prime }\}$
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Statements:
$[G^{\prime }:G]=p_1$
$⟹$
$G\in \{\text{ the normal subgroups of }{G}^{\prime }\}$
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2: Note

$|G^{\prime }|=p_1^{n_1}...p_k^{n_k}$ is not any pushed-on condition: the order of any group can be factorized into prime numbers like that, while of course $(p_1,...,p_k)$ and $(n_1,...,n_k)$ depend on $G^{\prime }$.

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3: Proof

Whole Strategy: Step 1: take the set of the conjugates of $G$, $C$, and take the group homomorphism, $f:G\to S_C$, such that $Ker(f)\in \{\text{ the subgroups of }{N}_{G^{\prime }}(G)\}$; Step 2: suppose that $G$ was not any normal subgroup hereafter, and see that $N_{G^{\prime }}(G)=G$; Step 3: suppose that $Ker(f)=G$, and find a contradiction; Step 4: suppose that $Ker(f)\in \{\text{ the proper subgroups of }G\}$, and find a contradiction; Step 5: conclude that $G$ is a normal subgroup of $G^{\prime }$.

Step 1:

Let us take the set of the conjugates of $G$, $C$.

There is the group homomorphism, $f:G^{\prime }\to S_C$, such that $Ker(f)\in \{\text{ the subgroups of }{N}_{G^{\prime }}(G)\}$, where $S_C$ is the permutations group of $C$ and $N_{G^{\prime }}(G)$ is the normalizer of $G$ on $G^{\prime }$, by the proposition that for any group, any subgroup, and the set of the conjugate of the subgroup, there is a group homomorphism from the group into the permutations group of the conjugates set, whose (the homomorphism's) kernel is a subgroup of the normalizer of the subgroup on the group.

Step 2:

Let us suppose that $G$ was not any normal subgroup hereafter.

Let us see that $N_{G^{\prime }}(G)=G$.

$N_{G^{\prime }}(G)$ contains $G$ and is a subgroup of $G^{\prime }$. If $N_{G^{\prime }}(G)=G^{\prime }$, $G$ would be a normal subgroup of $G^{\prime }$, a contradiction. So, $N_{G^{\prime }}(G)$ is a proper subgroup of $G^{\prime }$.

$[G^{\prime }:G]=p_1$ implies that $|G|=p_1^{n_1-1}...p_k^{n_k}$. Then, $|N_{G^{\prime }}(G)|=p_1^{n_1-1}...p_k^{n_k}$, by Lagrange's theorem: as $G$ is a subgroup of $N_{G^{\prime }}(G)$, $|N_{G^{\prime }}(G)|$ is a multiple of $|G|$, but as $N_{G^{\prime }}(G)$ is a subgroup of $G^{\prime }$, $|N_{G^{\prime }}(G)|$ is a divisor of $|G^{\prime }|$, so, $|N_{G^{\prime }}(G)|$ is a multiple of $p_1^{n_1-1}...p_k^{n_k}$ and is a divisor of $p_1^{n_1}...p_k^{n_k}$, but is not equal to $p_1^{n_1}...p_k^{n_k}$.

That means that $N_{G^{\prime }}(G)=G$.

So, $Ker(f)\in \{\text{ the subgroups of }{N}_{G^{\prime }}(G)=G\}$.

Step 3:

Let us suppose that $Ker(f)=G$.

As $G$ was not any normal subgroup of $G^{\prime }$, there would be a $g^{\prime }\in G^{\prime }$ such that $g^{\prime }G{g^{\prime }}^{-1}\ne G$, which would mean that there was a $g\in G$ such that $g^{\prime }g{g^{\prime }}^{-1}\notin G$.

$f(g^{\prime }g{g^{\prime }}^{-1})=f(g^{\prime })f(g)f({g^{\prime }}^{-1})=f(g^{\prime })1f({g^{\prime }}^{-1})$, because $Ker(f)=G$, $=f(g^{\prime }{g^{\prime }}^{-1})=f(1)=1$.

So, $g^{\prime }g{g^{\prime }}^{-1}\in Ker(f)$, a contradiction against $g^{\prime }g{g^{\prime }}^{-1}\notin G=Ker(f)$.

Step 4:

So, let us suppose that $Ker(f)$ was a proper subgroup of $G$.

$Ker(f)$ was a subgroup of $G^{\prime }$, by the proposition that for any group homomorphism, the quotient group of the domain by the kernel of the homomorphism is 'groups - homomorphisms' isomorphic to the range of the homomorphism: the 1st isomorphism theorem for groups.

So, $|Ker(f)|=p_1^{n_1^{\prime }}...p_k^{n_k^{\prime }}$, where $n_k^{\prime }\in \mathbb{N}$ such that $n_j^{\prime }\le n_j$, by Lagrange's theorem.

$|G^{\prime }/Ker(f)|=p_1^{n_1-n_1^{\prime }}...p_k^{n_k-n_k^{\prime }}$.

$l:=|C|\le p_1$, by the immediate corollary mentioned in Note for the proposition that for any group, any subgroup, and the left or right coset of the subgroup by any element of the group, the conjugates of the subgroup by any elements or the inverses of any elements respectively of the coset are the same.

$|S_C|=l!$.

$Ran(f)$ would be a subgroup of $S_C$, by the proposition that for any group homomorphism, the quotient group of the domain by the kernel of the homomorphism is 'groups - homomorphisms' isomorphic to the range of the homomorphism: the 1st isomorphism theorem for groups.

So, $|Ran(f)|$ would be a divisor of $l!$, by Lagrange's theorem.

By the proposition that for any group homomorphism, the quotient group of the domain by the kernel of the homomorphism is 'groups - homomorphisms' isomorphic to the range of the homomorphism: the 1st isomorphism theorem for groups, $|G^{\prime }/Ker(f)|=p_1^{n_1-n_1^{\prime }}...p_k^{n_k-n_k^{\prime }}=|Ran(f)|$.

As $|Ran(f)|$ was a divisor of $l!$ and $l\le p_1$, $|Ran(f)|=p_1=l=|G^{\prime }/Ker(f)|$: it could not be $1$, because if $|G^{\prime }/Ker(f)|=1$, $Ker(f)=G^{\prime }$, a contradiction against $Ker(f)\subset G\subset G^{\prime }$.

As $|G^{\prime }/G|=p_1$, $|G^{\prime }/Ker(f)|=|G^{\prime }/G|$, which would imply that $|Ker(f)|=|G|$, which would imply that $Ker(f)=G$, because $Ker(f)$ was a subgroup of $G$, but that would be a contradiction against that $Ker(f)$ was a proper subgroup of $G$.

Step 5:

So, supposing that $G$ was not any normal subgroup leads to a contradiction, so, $G$ is a normal subgroup of $G^{\prime }$.

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References

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