Prove that a group of order 42 has a nontrivial normal subgroup

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Homework Help Overview

The problem involves proving that a group of order 42 has a nontrivial normal subgroup, specifically utilizing Cauchy's Theorem while avoiding Sylow's Theorems.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of Cauchy's Theorem, noting the existence of an element of order 7. There is an exploration of the properties of the subgroup generated by this element and the conditions under which it may be normal. Questions arise regarding the intersection of two conjugate subgroups and the implications of their orders.

Discussion Status

The discussion is actively exploring the relationship between the subgroup generated by the element of order 7 and its conjugates. Participants are analyzing the consequences of their intersection and the implications for normality, with some guidance on using Lagrange's Theorem to assess subgroup orders.

Contextual Notes

Participants are working under the constraint of not using Sylow's Theorems, which influences their reasoning and approach to the problem.

Viking85
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Homework Statement



Prove that a group of order 42 has a nontrivial normal subgroup

Homework Equations



We are supposed to use Cauchys Theorem to solve the problem
We are not allowed to use any of Sylows Theorems


The Attempt at a Solution



By using Cauchys Theorem i know there there is an element of order 7
I then let A = <a> where o(a)=7
Essentially i know the normal subgroup is of order 7, but i need to still show it is normal
 
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If A isn't normal then it has a conjugate subgroup B which also has seven elements. What's the order of AB?
 
Well, I think that A ∩ B = 1 or 7 (Using Lagranges Theorem)
If A ∩ B is 1 it results in a contradiction because that would imply ord(AB)=49 > ord(G)

Which means that A ∩ B = 7

Which means the A = B ??
 
Viking85 said:
Well, I think that A ∩ B = 1 or 7 (Using Lagranges Theorem)
If A ∩ B is 1 it results in a contradiction because that would imply ord(AB)=49 > ord(G)

Which means that A ∩ B = 7

Which means the A = B ??

Right. That makes A normal, yes?
 

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