2024-06-30

661: 6-Elements Group Cannot Have 2 3-Elements Subgroups That Share Only Identity

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description/proof of that 6-elements group cannot have 2 3-elements subgroups that share only identity

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 any 6-elements group cannot have any 2 3-elements subgroups that share only the identity.

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 }
G1: { the subgroups of G}
G2: { the subgroups of G}
//

Statements:
(
|G|=6

|G1|=3

G1G2={1}
)

|G2|3
//


2: Natural Language Description


Any 6-elements group, G, cannot have any 2 3-elements subgroups, G1,G2, such that G1G2={1}.


3: Proof


Let us suppose that there were G1={1,a1,a2} and G2={1,b1,b2} such that G1G2={1}.

{1,a1,a2,b1,b2} would be distinct, and there would be the leftover, cG.

What would be b1a1? It could not be 1, because b1a1=1 would imply b1=a11G1G2={1}, which would imply b1=1, a contradiction. It could not be aj, because b1a1=aj would imply b1=aja11G1G2={1}, which would imply b1=1, a contradiction. It could not be bj, because b1a1=bj would imply a1=b11bjG1G2={1}, which would imply a1=1, a contradiction. So, b1a1=c. But likewise, b1a2=c, then, a1=b11c=a2, a contradiction. So, b1a1 cannot have any valid value.

So, the supposition is false.


References


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