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
- 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.
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:
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Statements:
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2: Note
3: Proof
Whole Strategy: Step 1: take the set of the conjugates of
Step 1:
Let us take the set of the conjugates of
There is the group homomorphism,
Step 2:
Let us suppose that
Let us see that
That means that
So,
Step 3:
Let us suppose that
As
So,
Step 4:
So, let us suppose that
So,
So,
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,
As
As
Step 5:
So, supposing that