2025-01-12

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

Topics


About: group

The table of contents of this article


Starting Context



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.

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.


Main Body


1: Structured Description


Here is the rules of Structured Description.

Entities:
G: { the groups }
|G|: =p1n1...pknk, where pj{ the prime numbers } and njN{0} such that p1<...<pk
G: { the subgroups of G}
//

Statements:
[G:G]=p1

G{ the normal subgroups of G}
//


2: Note


|G|=p1n1...pknk is not any pushed-on condition: the order of any group can be factorized into prime numbers like that, while of course (p1,...,pk) and (n1,...,nk) depend on G.


3: Proof


Whole Strategy: Step 1: take the set of the conjugates of G, C, and take the group homomorphism, f:GSC, such that Ker(f){ the subgroups of NG(G)}; Step 2: suppose that G was not any normal subgroup hereafter, and see that NG(G)=G; Step 3: suppose that Ker(f)=G, and find a contradiction; Step 4: suppose that Ker(f){ the proper subgroups of G}, and find a contradiction; Step 5: conclude that G is a normal subgroup of G.

Step 1:

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

There is the group homomorphism, f:GSC, such that Ker(f){ the subgroups of NG(G)}, where SC is the permutations group of C and NG(G) is the normalizer of G on G, 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 NG(G)=G.

NG(G) contains G and is a subgroup of G. If NG(G)=G, G would be a normal subgroup of G, a contradiction. So, NG(G) is a proper subgroup of G.

[G:G]=p1 implies that |G|=p1n11...pknk. Then, |NG(G)|=p1n11...pknk, by Lagrange's theorem: as G is a subgroup of NG(G), |NG(G)| is a multiple of |G|, but as NG(G) is a subgroup of G, |NG(G)| is a divisor of |G|, so, |NG(G)| is a multiple of p1n11...pknk and is a divisor of p1n1...pknk, but is not equal to p1n1...pknk.

That means that NG(G)=G.

So, Ker(f){ the subgroups of NG(G)=G}.

Step 3:

Let us suppose that Ker(f)=G.

As G was not any normal subgroup of G, there would be a gG such that gGg1G, which would mean that there was a gG such that ggg1G.

f(ggg1)=f(g)f(g)f(g1)=f(g)1f(g1), because Ker(f)=G, =f(gg1)=f(1)=1.

So, ggg1Ker(f), a contradiction against ggg1G=Ker(f).

Step 4:

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

Ker(f) was a subgroup of G, 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)|=p1n1...pknk, where nkN such that njnj, by Lagrange's theorem.

|G/Ker(f)|=p1n1n1...pknknk.

l:=|C|p1, 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.

|SC|=l!.

Ran(f) would be a subgroup of SC, 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/Ker(f)|=p1n1n1...pknknk=|Ran(f)|.

As |Ran(f)| was a divisor of l! and lp1, |Ran(f)|=p1=l=|G/Ker(f)|: it could not be 1, because if |G/Ker(f)|=1, Ker(f)=G, a contradiction against Ker(f)GG.

As |G/G|=p1, |G/Ker(f)|=|G/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.


References


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